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# Ncert Solutions For Class 12 Maths Ex 9.5

## Ncert Solutions For Class 12 Maths Chapter 9 Ex 9.5

Q.1: (x2 + xy)dy = (x2 + y2)dx

Ans:

Given:

(x2 + xy)dy = (x2 + y2)dx

$$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}$$ . . . . . . . . . . . (1)

Let, F(x, y) = $$\frac{x^{2}+y^{2}}{x^{2}+xy}$$

Now,

F(λx, λy) = $$\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x )(\lambda y)}=\frac{x^{2}+y^{2}}{x^{2}+xy}=\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{x^{2}+(vx)^{2}}{x^{2}+x(vx)}\\ \\ \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^{2}}{1+v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{1+v}-v=\frac{(1+v^{2})-v(1+v)}{1+v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{1-v}{1+v}\\ \\ \Rightarrow (\frac{1+v}{1-v})=dv=\frac{dx}{x}\\ \\ \Rightarrow (\frac{2-1+v}{1-v})dv=\frac{dx}{x}\\ \\ \Rightarrow (\frac{2}{1-v}-1)dv=\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ -2log(1-v)-v=log x – log k

$$\Rightarrow v=-2log(1-v)-log x+log k\\ \\ \Rightarrow v=log[\frac{k}{x(1-v)^{2}}]\\ \\ \Rightarrow \frac{y}{x}=log[\frac{k}{x(1-\frac{y}{x})^{2}}]\\ \\ \Rightarrow \frac{y}{x}=log[\frac{kx}{(x-y)^{2}}]\\ \\ \Rightarrow \frac{kx}{(x-y)^{2}}=e^{\frac{y}{x}}\\ \\ \Rightarrow (x-y)^{2}=k\;x\;e^{-\frac{y}{x}}\\$$

This is the required solution of the given differential equation.

Q.2: y = $$\frac{x+y}{x}$$

Ans:

Given:

y = $$\frac{x+y}{x}$$

$$\Rightarrow \frac{dy}{dx}=\frac{x+y}{x}$$ . . . . . . . . . . (1)

Let F(x,y)= $$\frac{x+y}{x}$$

Now, F(λx,λy)= $$\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=\lambda ^{0}F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{x+vx}{x}\\ \\ \Rightarrow v+x\frac{dv}{dx}=1+v\\ \\ \Rightarrow x\frac{dv}{dx}=1\\ \\ \Rightarrow dv=\frac{dx}{x}$$

Integrate on both the sides, we get:

V = log x + C

$$\Rightarrow \frac{y}{x}=logx+c\\ \\ \Rightarrow y=xlogx+Cx$$

This is the required solution of the given differential equation.

Q.3: (x-y)dy-(x+y)dx=0

Ans:

Given:

(x – y)dy – (x + y)dx = 0

$$\Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}$$ . . . . . . . . . . (1)

Let, F(x, y) = $$\frac{x+y}{x-y}$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{x+y}{x-y}=\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}\\ \\ x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{1-v}\\ \\ \Rightarrow \frac{1-v}{(1+v^{2})}dv=\frac{dx}{x}\\ \\ \Rightarrow (\frac{1}{1+v^{2}}-\frac{v}{1-v^{2}})dv=\frac{dx}{x}\\$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$tan^{-1}v-\frac{1}{2}log(1+v^{2})=log\, x+C\\ \\ \Rightarrow tan^{-1}(\frac{y}{x})-\frac{1}{2}log[1+(\frac{y}{x})^{2}]=log\, x+C\\ \\ \Rightarrow tan^{-1}(\frac{y}{x})-\frac{1}{2}log(\frac{x^{2}+y^{2}}{x^{2}})=log\, x+C\\ \\ \Rightarrow tan^{-1}(\frac{y}{x})-\frac{1}{2}[log(x^{2}+y^{2})-log\, x^{2}]=log\, x+C\\ \\ \Rightarrow tan^{-1}(\frac{y}{x})=\frac{1}{2}log(x^{2}+y^{2})+C\\$$

This is the required solution of the given differential equation.

Q.4: (x2 – y2) dx + 2xy dy = 0

Ans:

Given,

(x2 – y2)dx + 2xy dy = 0

$$\Rightarrow \frac{dy}{dx}=\frac{-(x^{2}-y^{2})}{2xy}$$ . . . . . . . . . (1)

Let, F(x, y) = $$\frac{-(x^{2}-y^{2})}{2xy}$$

Therefore, $$F(\lambda x,\lambda y)=[\frac{(\lambda x)^{2}-(\lambda y)^{2}}{2xy}]=\frac{-(x^{2}-y^{2})}{2(\lambda x)(\lambda y)}=\frac{-(x^{2}-y^{2})}{2xy}=\lambda^{\circ}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=-[\frac{x^{2}-(vx)^{2}}{2x\cdot (vx)}]=\frac{v^{2}-1}{2v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-1}{2v}-v=\frac{v^{2}-1-2v^{2}}{2v}\\ \\ \Rightarrow x\frac{dv}{dx}=-\frac{(1+v^{2})}{2v}\\ \\ \Rightarrow \frac{2v}{1+v^{2}}dv=-\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ Log(1 + v2) = -log x + log C = $$log\frac{C}{x}$$

$$\Rightarrow 1+v^{2}=\frac{C}{x}\\ \\ \Rightarrow [1+\frac{y^{2}}{x^{2}}]=\frac{C}{x}\\ \\ \Rightarrow x^{2}+y^{2}=Cx$$

This is the required solution of the given differential equation.

Q.5: $$x^{2}\frac{dy}{dx}-x^{2}-2y^{2}+xy$$

Ans:

Given:

$$x^{2}\frac{dy}{dx}-x^{2}-2y^{2}+xy$$

$$\frac{dy}{dx}=\frac{x^{2}-2y^{2}+xy}{x^{2}}$$ . . . . . . . . . . . (1)

Let F(x,y)= $$\frac{ x^{2}- 2y^{2} + xy}{ x^{2}}$$

$$F( \lambda x, \lambda y)= \frac{( \lambda x )^{ 2}-2( \lambda y)^{2}+( \lambda x)( \lambda x)}{( \lambda x)^{2}}=\frac{ x^{2} – 2y^{2}+ x y }{ x^{2}}=\lambda ^{\circ}\cdot F( x , y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{ dy }{ dx } = v + x \frac { dv }{ dx }$$

Substitute the values of v and $$\frac{ dy }{ dx }$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{x^{2}-2(vx)^{2}+x\cdot (vx)}{x^{2}}\\ \\ \Rightarrow v+x\frac{dv}{dx}=1-2v^{2}+v\\ \\ \Rightarrow x\frac{dv}{dx}=1-2v^{2}\\ \\ \Rightarrow \frac{dv}{1-2v^{2}}=\frac{dx}{x}\\ \\ \Rightarrow \frac{1}{2}\cdot \frac{dv}{\frac{1}{2}-v^{2}}=\frac{dx}{x}\\ \\ \Rightarrow \frac{1}{2}[\frac{dv}{(\frac{1}{\sqrt{2}})^{2}-v^{2}}]=\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$\frac {1} {2} \cdot \frac {1} { 2\times \frac{1} {\sqrt {2} }}log\left | \frac{ \frac {1} { \sqrt {2} }+v}{ \frac {1} {\ sqrt {2}}-v} \right |=log\left | x \right |+C\\ \\ \\ \Rightarrow \frac {1} {2\sqrt {2}}log\left | \frac{\frac {1} {\sqrt {2}}+\frac {y}{x}}{\frac {1}{ \sqrt {2}}-\frac {y} {x} } \right |=log\left | x \right |+C\\ \\ \\ \Rightarrow \frac {1} {2\times \frac {1} {\sqrt {2}}}log\left | \frac{x + 2\sqrt {2} y}{x + 2 \sqrt {2} y} \right |=log\left | x \right |+C$$

This is the required solution of the given differential equation.

Q.6: xdy – ydx = $$\sqrt{x^{2}+y^{2}}dx$$

Ans:

xdy – ydx = $$\sqrt{x^{2}+y^{2}}dx$$

$$\Rightarrow xdy=[y+\sqrt{x^{2}+y^{2}}]dx\\ \\ \Rightarrow \frac{dy}{dx} =\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$$ . . . . . . . . . . . . . (1)

Let, F(x,y) = $$\frac{y+\sqrt{x^{2}+y^{2}}}{x^{2}}$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}(\lambda y)^{2}}}{x}=\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^{2}+(vx)^{2}}}{x}\\ \\ \Rightarrow v+x\frac{dv}{dx}=v+\sqrt{1+v^{2}}\\ \\ \Rightarrow \frac{dv}{\sqrt{1+v^{2}}}=\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$log\left | v+\sqrt{1+v^{2}} \right |=log\left | x \right |+logC\\ \\ \Rightarrow log\left | \frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}} \right |=log\left | Cx \right |\\ \\ \Rightarrow log\left | \frac{y+\sqrt{x^{2}+y^{2}}}{x} \right |=log\left | Cx \right |\\ \\ \Rightarrow y+\sqrt{x^{2}+y^{2}}=Cx^{2}$$

This is the required solution of the given differential equation.

Q.7: $$\left \{xcos(\frac{y}{x})+ysin(\frac{y}{x})\right \} ydx =\left \{ ysin(\frac{y}{x} ) -xcos(\frac{y}{x})\right \}x\;dy$$

Ans:

Given:

$$\left \{xcos( \frac {y}{x})+y sin( \frac {y} {x})\right \} y dx =\left \{ ysin( \frac {y}{x} ) -xcos( \frac {y}{x})\right \} xdy\\$$

$$\\\frac{dy}{dx}=\frac{\left \{xcos(\frac{y}{x})+ysin(\frac{y}{x})\right \} y }{\left \{ ysin(\frac{y}{x} ) -xcos(\frac{y}{x})\right \}x}$$ . . . . . . . . . . (1)

Let, F(x, y) = $$\frac{dy}{dx}=\frac{\left \{xcos(\frac{y}{x})+ysin(\frac{y}{x})\right \} y }{\left \{ ysin(\frac{y}{x} ) -xcos(\frac{y}{x})\right \}x}$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\left \{\lambda xcos(\frac{\lambda y}{\lambda x})+\lambda ysin(\frac{\lambda y}{\lambda x})\right \} \lambda y }{\left \{ \lambda ysin(\frac{\lambda y}{\lambda x} ) -\lambda xcos(\frac{\lambda y}{\lambda x})\right \}\lambda x}\\ \\ =\frac{\left \{xcos(\frac{y}{x})+ysin(\frac{y}{x})\right \} y }{\left \{ ysin(\frac{y}{x} ) -xcos(\frac{y}{x})\right \}x}\\ \\ =\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

$$\boldsymbol{\Rightarrow }$$ Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$v+x\frac{dv}{dx}=\frac{(x\,cos\,v+vx\,sin\,v)\cdot vx}{(vx\,sin\,v-x\,cos\,v)\cdot x}\\ \\ \Rightarrow v+x\frac{dv}{dx}=\frac{v\,cos\,v+v^{2}\,sin\,v}{v\,sin\,v-cos\,v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{v\,cos\,v+v^{2}\,sin\,v}{v\,sin\,v-cos\,v}-v\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{v\,cos\,v+v^{2}\,sin\,v-v^{2}\,sin\,v+v\,cos\,v}{v\,sin\,v-cos\,v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{2v\,cos\,v}{v\,sin\,v-cos\,v}\\ \\ \Rightarrow [\frac{v\,sin\,v-cos\,v}{v\,cos\,v}]dv=\frac{2dx}{x}\\ \\ \Rightarrow (tan\,v-\frac{1}{v})dv=\frac{2dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ Log(sec v) – log v = 2 log x + log C

$$\Rightarrow log ( \frac { sec \,v}{ v })=log\,(Cx^{2})\\ \\ \Rightarrow log( \frac{ sec \,v}{v})=Cx^{2}\\ \\ \Rightarrow \sec \,v = Cx^{2}v\\ \\ \Rightarrow sec(\frac{y}{x})= C-x^{2} \cdot \frac{y}{x}\\ \\ \Rightarrow sec(\frac{y}{x})=Cxy\\ \\ \Rightarrow sec(\frac{y}{x})=\frac{1}{Cxy}=\frac{1}{C}\cdot \frac{1}{xy}\\ \\ \Rightarrow xy\,cos(\frac{y}{x})=k\: \: \: \: \: \: \: \: \: \: \: \: (k=)\frac{1}{C}$$

This is the required solution of the given differential equation.

Q.8: $$x\frac{dy}{dx}-y+x\,sin(\frac{y}{x})=0$$

Ans:

$$x\frac{dy}{dx}-y+x\,sin(\frac{y}{x})=0$$

$$\Rightarrow x\frac{dy}{dx}=y-xsin(\frac{y}{x})\\ \\ \Rightarrow \frac{dy}{dx}=\frac{y-xsin(\frac{y}{x})}{x}$$ . . . . . . . . . (1)

Let, F(x , y) = $$\frac{y-xsin(\frac{y}{x})}{x}$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\lambda y-\lambda xsin(\frac{\lambda y}{\lambda x})}{\lambda x}=\frac{y-xsin(\frac{y}{x})}{x}=\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{vx-x\,sin\,v}{x}\\ \\ \Rightarrow v+x\frac{dv}{dx}=v-sin\, v\\ \\ \Rightarrow -\frac{dv}{sin\, v}=-\frac{dx}{x}\\ \\ \Rightarrow cosec\,v\,dv=-\frac{dx}{x}\\$$

Integrate on both the sides:

$$\boldsymbol{\Rightarrow }$$ $$log\left |cosec v – cot v \right |=-logx+log C=log\frac{C}{x}\\ \\ \Rightarrow cosec(\frac{y}{x})-cot(\frac{y}{x})=\frac{C}{x}\\ \\ \Rightarrow \frac{1}{sin(\frac{y}{x})}-\frac{cos(\frac{y}{x})}{sin(\frac{y}{x})}=\frac{C}{x}\\ \\ \Rightarrow x[1-cos(\frac{y}{x})]=Csin(\frac{y}{x})\\$$

This is the required solution of the given differential equation.

Q.9: y dx + x log ($$\frac{y}{x}$$) dy – 2x dy = 0

Ans:

$$ydx + xlog(\frac{y}{x})dy-2xdy=0\\ \\ \Rightarrow ydx =[2x-xlog(\frac{y}{x})]dy\\ \\ \Rightarrow \frac{dy}{dx}=\frac{y}{2x-xlog(\frac{y}{x})}$$ . . . . . . . . . (1)

Let, F(x, y) = $$\frac{y}{2x-xlog(\frac{y}{x})}$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\lambda y}{2\lambda x-\lambda xlog(\frac{\lambda y}{\lambda x})}=\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let y = vx

Differentiate both the sides w.r.t. x, we get:

$$\frac{dy}{dx}=v+x\frac{dv}{dx}\\$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=\frac{vx}{2x-xlogv}\\ \\ \Rightarrow v+x\frac{dv}{dx}=\frac{v}{2-logv}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{v}{2-logv}-v\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{v-2v+vlogv}{2-logv}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{vlogv-v}{2-logv}\\ \\ \Rightarrow \frac{2-logv}{vlogv-v}dv=\frac{dx}{x}\\ \\ \Rightarrow [\frac{1+(1-logv)}{v(logv-1)}]dv=\frac{dx}{x}\\ \\ \Rightarrow [\frac{1}{v(logv-1)}-\frac{1}{v}]dv=\frac{dx}{x}\\$$

Integrate on both the sides:

$$\int \frac{1}{v(logv-1)}dv-\int \frac{1}{v}dv=\int \frac{1}{x}dx\\ \\ \Rightarrow \int \frac{1}{v(logv-1)}dv-log v =logx+logC$$ . . . . . . . . . . . . (2)

Let, log v – 1 =t

$$\Rightarrow \frac{d}{dv}(log v-1)=\frac{dt}{dv}\\ \\ \Rightarrow \frac{1}{v}=\frac{dt}{dv}\\ \\ \Rightarrow \frac{dv}{v}=dt$$

So, equation (1) will become:

$$\int \frac{dt}{t}-logv=logx+logC\\ \\ \Rightarrow log t-log(\frac{y}{x})=log(Cx)\\ \\ \Rightarrow log[log(\frac{y}{x})-1]-log(\frac{y}{x})=log(Cx)\\ \\ \Rightarrow log[\frac{log(\frac{y}{x})-1}{ \frac{y}{x}}]=log(Cx)\\ \\ \Rightarrow \frac{x}{y}[log(\frac{y}{x})-1]=Cx\\ \\ \Rightarrow log(\frac{y}{x})-1=Cy$$

This is the required solution of the given differential equation.

Q.10: $$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$$

Ans:

$$\boldsymbol{\Rightarrow }$$ $$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\\ \\ \Rightarrow (1+e^{\frac{x}{y}})dx=-e^{\frac{x}{y}}(1-\frac{x}{y})dy\\ \\ \Rightarrow \frac{dx}{dy}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$$ . . . . . . . . . . (1)

Let F(x,y) = $$\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}$$

Therefore, $$F(\lambda x,\lambda y)= \frac{-e^{\frac{\lambda x}{\lambda y}}(1-\frac{\lambda x}{\lambda y})}{1+e^{\frac{\lambda x}{\lambda y}}}\\ \\ =\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}\\ \\ =\lambda ^{0}\cdot F(x,y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, x = vy

$$\frac{d}{dy}(x)=\frac{d}{dy}(vy)\\ \\ \Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$$

Differentiate both the sides w.r.t. x, we get

Substitute the values of v and $$\frac{dx}{dy}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v + y \frac {dv}{dx} =\frac{ -e^{v}(1-v)}{1+e}\\ \\ \Rightarrow y\frac{dv}{dy} = \frac{-e^{v} + ve^{v}}{1+e^{v}}-v\\ \\ \Rightarrow y\frac{dv}{dy} = \frac{-e^{v}+ve^{v}-v-ve^{v}}{1 + e^{v}}\\ \\ \Rightarrow y\frac{dv}{dy} = -[\frac{v+e^{v}}{1 + e^{v}}]\\ \\ \Rightarrow [\frac{v+e^{v}}{1+e^{v}}]dv = -\frac{dy}{y}$$

Integrate on both the sides, we get:

$$log(v+e^{v})=-logy+logC=log(\frac{C}{y})\\ \\ \Rightarrow [\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}\\ \\ \Rightarrow x+ye^{\frac{x}{y}}=C\\$$

This is the required solution of the given differential equation.

Q.11: ( x + y )dy + ( x – y )dx = 0 ; y = 1 when x = 1

Ans:

(x + y)dy + (x – y)dx = 0

$$\boldsymbol{\Rightarrow }$$ ( x + y )dy = -( x – y )dx

$$\boldsymbol{\Rightarrow }$$ $$\frac{dy}{dx}=\frac{-(x-y)}{x+y}$$ . . . . . . . . . . . . (1)

Let, F( x , y ) = $$\frac{-(x-y)}{x+y}$$

Therefore, $$F( \lambda x,\lambda y) = \frac{-(\lambda x – \lambda y)}{\lambda x + \lambda y} = \frac{-( x- y)}{ x + y} =\lambda ^{0}\cdot F(x , y)$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y=vx

Differentiate both the sides w.r.t. x, we get:

$$\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)$$ $$\frac{dy}{dx}=v+x\frac{dv}{dx}\\$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx} = \frac{-(x-vx)}{x+vx}\\ \\ \Rightarrow v+x\frac{dv}{dx} = \frac{v-1}{v+1}\\ \\ x\frac{dv}{dx} = \frac{v-1}{v+1}-v\\ \\ x\frac{dv}{dx} = \frac{v-1-v(v+1)}{v+1}\\ \\ x\frac{dv}{dx} = \frac{v-1-v^{2}-v}{v+1} = \frac{-(1+v^{2})}{v+1}\\ \\ \Rightarrow \frac{(v+1)}{1+v^{2}}dv = -\frac{dx}{x}\\ \\ \Rightarrow [\frac{v}{1+v^{2}}+\frac{1}{1+v^{2}}]dv = -\frac{dx}{x}\\$$

Integrate on bothe the sides, we get:

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}log(1 + v^{2}) + tan^{-1}v = -logx+k\\ \\ \Rightarrow log(1 + v^{2}) + 2tan^{-1}v = -2logx+2k\\ \\ \Rightarrow log[(1 + v^{2})\cdot x^{2}] + 2tan^{-1}v = 2k\\ \\ \Rightarrow log[(1 + \frac{y^{2}}{x^{2}})\cdot x^{2}] + 2tan^{-1}\frac{y}{x} = 2k\\ \\ \Rightarrow log(x^{2} + y^{2}) + 2tan^{-1}\frac{y}{x} = 2k$$ …..(2)

Now y = 1 at x = 1:

$$\Rightarrow log 2 + 2tan^{-1}1 = 2k\\ \\ \Rightarrow log2 + 2\times \frac{\pi }{4} = 2k\\ \\ \Rightarrow \frac{\pi }{2} + log2 = 2k\\$$

Substitute value of 2k in equn(2), we get:

$$log(x^{2} + y^{2}) + 2tan^{-1}(\frac{y}{x}) = \frac{\pi }{2} + log2$$

This is the required solution of the given differential equation.

Q.12: x2 dy + ( xy + y2 )dx = 0, y = 1 when x =1

Ans:

x2 dy + ( xy + y2 )dx = 0

$$\Rightarrow x^{2}dy = -(xy+y^{2})dx\\ \\ \Rightarrow \frac{dy}{dx} = \frac{-(xy+y^{2})}{x^{2}}$$ . . . . . . . . . . . . (1)

Let F(x, y) = $$\frac{-(xy+y^{2})}{x^{2}}$$

Therefore, $$F( \lambda x , \lambda y ) \frac{ -( \lambda x \cdot \lambda y + (\lambda y)^{2})}{(\lambda x)^{2}} = \frac{-(xy+y^{2})}{x^{2}} = \lambda ^{0}\cdot F (x , y )$$

Here we have observed that equation (1) is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get:

$$\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)$$ $$\frac{dy}{dx}=v+x\frac{dv}{dx}$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx} = \frac{ -[x\cdot vx + ( vx )^{2}]}{ x^{2}} = -v-v^{2}\\ \\ \Rightarrow x\frac{dv}{dx} = -v^{2}-2v = -v( v + 2 )\\ \\ \Rightarrow \frac{dv}{v ( v + 2 )} = -\frac{dx}{x}\\ \\ \Rightarrow \frac{1}{2}[\frac{( v + 2 ) – v}{v( v + 2 )}]dv = -\frac{dx}{x}\\ \\ \Rightarrow \frac{1}{2}[\frac{1}{v} – \frac{1}{v + 2}]dv = -\frac{dx}{x}\\$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}[logv – log( v + 2 )] = -logx + logC\\ \\ \Rightarrow \frac{1}{2}log(\frac{v}{v+2}) = log\frac{C}{x}\\ \\ \Rightarrow \frac{v}{v + 2} = (\frac{C}{x})^{2}\\ \\ \Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2} = (\frac{C}{x})^{2}\\ \\ \Rightarrow \frac{y}{y + 2x} = \frac{C^{2}}{x^{2}}\\ \\ \frac{x^{2}y}{y+2x} = C^{2}$$ . . . . . . . . . . . . (2)

Now, y = 1 at x = 1:

$$\Rightarrow \frac{1}{1 + 2} = C^{2}\\ \\ \Rightarrow C^{2} = \frac{1}{3}$$

Substituting C2 = $$\frac{1}{3}$$

$$\frac{x^{2}y}{y + 2x} = \frac{1}{3}\\ \\ \Rightarrow y + 2x = 3x^{2}y$$

This is the required solution for the given differential equation.

Q.13: $$[xsin^{2}(\frac{x}{y}-y)]dx+xdy=0;\, y=\frac{\pi }{4}\, when\, x=1$$

Ans:

$$[xsin^{2}(\frac{x}{y}-y)]dx+xdy=0\\ \\ \Rightarrow \frac{dy}{dx}=\frac{-[xsin^{2}(\frac{y}{x})-y]}{x}\\ \\$$ . . . . . . . . . . . . . . (1)

Let, F(x , y) = $$\frac{-[xsin^{2}(\frac{y}{x})-y]}{x}$$

Therefore, $$F(\lambda x,\lambda y)\frac{-[\lambda x\cdot sin^{2}(\frac{\lambda x}{\lambda y})-\lambda y]}{\lambda x} = \frac{-[xsin^{2}(\frac{y}{x})-y]}{x} = \lambda ^{0}\cdot F(x,y)$$

So, the given differential equation is a homogeneous equation.

Let y = vx

Differentiate both the sides w.r.t. x, we get

$$\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)$$ $$\frac{dy}{dx}=v+x\frac{dv}{dx}\\$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx} = \frac{-[xsin^{2}v-vx]}{x}\\ \\ \Rightarrow v+x\frac{dv}{dx} = -[sin^{2}v-v] = v – sin^{2}v\\ \\ \Rightarrow x\frac{dv}{dx } = -sin^{2}v\\ \\ \Rightarrow \frac{dv}{sin^{2}v} = -\frac{dx}{x}\\ \\ \Rightarrow cosec^{2}vdv = -\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ -cot v = -log|x|- C

$$\Rightarrow cot v=log\left | x \right |+C\\ \\ \Rightarrow cot (\frac{y}{x})=log\left | x \right |+logC\\ \\ \Rightarrow cot (\frac{y}{x})=log\left |Cx \right |\\$$ . . . . . . . . . . . (2)

Now, y = $$\frac{\pi }{4}$$ at x = 1

$$\boldsymbol{\Rightarrow }$$ $$\cot\frac{\pi }{4} = log\left | C \right |$$

$$\boldsymbol{\Rightarrow }$$ 1 = log C

$$\boldsymbol{\Rightarrow }$$ C = e1 = e

Substituting C = e in equation (2), we get:

$$cot(\frac{y}{x})=log\left | ex \right |\\$$

This is the required solution for the given differential equation.

Q.14: $$\frac{dy}{dx}-\frac{y}{x}+cosec(\frac{y}{x})=0;y=0\:when\:x=1$$

Ans:

$$\frac{dy}{dx}-\frac{y}{x}+cosec(\frac{y}{x})=0\\ \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-cosec(\frac{y}{x})$$ . . . . . . . . . . . (1)

Let, F(x , y) = $$\frac{y}{x}-cosec(\frac{y}{x})$$

Therefore, $$F(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}-cosec(\frac{\lambda y}{\lambda x})$$

$$\boldsymbol{\Rightarrow }$$ $$F(\lambda x,\lambda y)=\frac{ y}{ x}-cosec(\frac{ y}{ x})=F(x,y)=\lambda ^{0}\cdot F(x,y)$$

So, the given differential equation is a homogeneous equation.

Let, y = vx

Differentiate both the sides w.r.t. x, we get

$$\Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)$$ $$\frac{dy}{dx}=v+x\frac{dv}{dx}\\$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v+x\frac{dv}{dx}=v-cosec\,v\\ \\ \Rightarrow -\frac{dv}{cosec\,v}=-\frac{dx}{x}\\ \\ \Rightarrow -sin\,vdv=\frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$cos\,v = log\,x+log\,C = log\left | Cx \right |\\ \\ \Rightarrow cos(\frac{y}{x}) = log\left | Cx \right |$$ . . . . . . . . . . . . . . . . . (2)

This is the required solution for the given differential equation.

Now, y = 0 at x = 1

$$\boldsymbol{\Rightarrow }$$ $$cos(0)=log\;C$$

$$\boldsymbol{\Rightarrow }$$ 1 = log C

$$\boldsymbol{\Rightarrow }$$ C = e1 = e

This is the required solution for the given differential equation.

Q.15: $$2xy+y^{2}-2x^{2}\frac{dy}{dx}=0;\,y=2\:when\:x=1$$

Ans:

$$2xy+y^{2}-2x^{2}\frac{dy}{dx}=0$$

$$\Rightarrow 2x^{2}\frac{dy}{dx}=2xy+y^{2}\\ \\ \Rightarrow \frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}}$$ . . . . . . . . . . . . (1)

Let, F( x , y ) = $$\frac{2xy+y^{2}}{2x^{2}}\\$$

Therefore, $$F(\lambda x,\lambda y)=\frac{2(\lambda x)(\lambda y)+(\lambda y)^{2}}{2(\lambda x)^{2}}=\frac{2xy+y^{2}}{2x^{2}}=\lambda ^{0}\cdot F(x,y)$$

So, the given differential equation is a homogeneous equation.

Let, y=vx

Differentiate both the sides w.r.t. x, we get:

$$\boldsymbol{\Rightarrow }$$ $$\frac{d}{dx}(y)=\frac{d}{dx}(vx)$$

$$\frac{dy}{dx}=v+x\frac{dv}{dx}\\$$

Substitute the values of v and $$\frac{dy}{dx}$$ in equation(1), we get:

$$\boldsymbol{\Rightarrow }$$ $$v + x\frac{dv}{dx} = \frac{2x(vx)_(vx)^{2}}{2x^{2}}\\ \\ \Rightarrow v + x\frac{dv}{dx} = \frac{2v+v^{2}}{2}\\ \\ \Rightarrow v+x\frac{dv}{dx} = v + \frac{v^{2}}{2}\\ \\ \Rightarrow \frac{2}{v^{2}}dv = \frac{dx}{x}$$

Integrate on both the sides, we get:

$$\boldsymbol{\Rightarrow }$$ $$2\cdot \frac{v^{-2+1}}{-2+1} = log\left | x \right |+C\\ \\ \Rightarrow -\frac{2}{v} = log\left | x \right |+C\\ \\ \Rightarrow -\frac{2}{\frac{y}{x}} = log\left | x \right |+C\\ \\ \Rightarrow -\frac{2x}{y} = log\left | x \right |+C$$ . . . . . . . . . . . (2)

Now, y = 2 at x = 1

$$\boldsymbol{\Rightarrow }$$ -1 = log(1)+C

$$\boldsymbol{\Rightarrow }$$ C = -1

Substutute C = -1 in equation (2), we get:

$$\boldsymbol{\Rightarrow }$$ $$-\frac{2x}{y}=log\left | x \right |-1\\ \\ \Rightarrow \frac{2x}{y}=1-log\left | x \right |\\ \\ \Rightarrow y=\frac{2x}{1-log\left | x \right |},(x\neq 0,x\neq e)\\$$

This is the required solution of the given differential equation.

Q.16: A homogeneous differential equation of the form $$\frac{dx}{dy}=h(\frac{x}{y})$$ can be solved by making the substitution

(i) y = vx

(ii) v = yx

(iii) x = vy

(iv) x = v

Ans:

For solving the homogeneous equation of the form $$\frac{dx}{dy}=h(\frac{x}{y})$$ , we need to make the substitution as x = vy. Hence, the correct answer is (iii).

Q.17: Which of the following is a homogeneous differential equation?

(i) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0

(ii) (xy)dx – (x3 + y3)dy = 0

(iii) (x3 + 2y2)dx + 2xy dy = 0

(iv) y2 dx + (x2 –xy2 – y2)dy = 0

Ans:

Function F(x, y) is said to be the homogenous function of degree n, if

F( λx , λy ) = λn F(x, y) for any non-zero constant (λ).

Consider the equation given in alternative IV:

Y2 dx + ( x2 – xy –y2 )dy = 0

$$\boldsymbol{\Rightarrow }$$ $$\frac{dy}{dx}=\frac{-y^{2}}{x^{2}-xy-y^{2}}=\frac{y^{2}}{y^{2}+xy-x^{2}}\\$$

Let F( x , y ) = $$\frac{y^{2}}{y^{2}+xy-x^{2}}$$

$$\boldsymbol{\Rightarrow }$$ $$F(\lambda x,\lambda y) = \frac{(\lambda y)^{2}}{(\lambda y)^{2}+(\lambda x)(\lambda y)-(\lambda x)^{2}} = \frac{\lambda ^{2}y^{2}}{\lambda ^{2}(y^{2}+xy-x^{2})}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\lambda ^{0}(\frac{y^{2}}{y^{2}+xy-x^{2}}) = \lambda ^{0}\cdot F(x,y)$$

Hence, the differential equation given in alternative (iv) is a homogenous equation.

format: plain


text:
Orbital period of an asteroid

Tags:
1. Feb 22, 2017

TheBigDig

1. The problem statement, all variables and given/known data
The orbit of an asteroid extends from the Earth’s orbit to Jupiter’s orbit, just touching both. Assume that the planetary orbits are circular and co-planar and that Newton’s constant G, the mass of the sun Ms, the mass of the asteroid ma and the radii of the Earth’s and Jupiter’s orbit respectively, (LE, LJ ), are known.
1. Find the orbital period of the asteroid.

Express all of your answers in terms of the given quantities of the problem: G, Ms, ma,LE and LJ

2. Relevant equations
T = 2π (a3/μ)1/2 where T is the period, a is the semi-major axis of the orbit and μ is the gravitational mass of the sun (μ = GMs)

3. The attempt at a solution
I'm actually really stuck on this one though it seems simple. It seems like the answer is T = 2π (LJ3/μ)1/2 but I am almost certain that is wrong. It just seems to easy to me.

Last edited by a moderator: Feb 22, 2017
2. Feb 22, 2017

Isaac0427

That is because for a you used the orbital radius of Jupiter's orbit. What you found was the period of Jupiter's orbit. What you need here is the semi-major axis. Try a diagram.

3. Feb 22, 2017

Staff: Mentor

Make a sketch of the orbit and identify the major axis.


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# Z Transform Formula

## What is Z Transform?

Last updated date: 26th Mar 2023
Total views: 192.9k
Views today: 2.70k

In order to define z transform, it is simply a mathematical mechanism of going from the discrete time domain to the z domain also referred to as the complex frequency domain. Under the discrete time domain, a signal is generally described as a series of real or complex numbers which is then converted to the z-domain by the mechanism of z-transform.

### Z Transform Formula

The formula applied in order to convert a discrete time signal x[n] to X[z] is as below:

X(Z)|z = e = F. T[x(n)]

Where,

x[n] represents Finite length signal

[0, N] represents Sequence support interval

z  represents Any complex number

N represents Integer

X[z] represents the z-transform of the discrete time signal.

Now here is to notify that the z-transform comes in two parts. The first part is the formula as given above while the second part is to describe a region of convergence (ROC) for the z-transform. Both parts are required for a complete z-transform as a z-transform unaccompanied by ROC would not be of much assistance in signal processing. Moreover, though the z-transform is acquired by directly using this formula, the inverse z-transform needs some mathematical manipulations which are linked to the power series and geometric series.

### Z Transformation Relation To Discrete-Time Fourier Transform (DTFT)

There is a close connection between DTFT and z-transform. In fact, each of their respective formulas are also pretty similar, which is most commonly overlooked. Thus, let’s take a look at the formulas for both the DTFT and the z-transform for a signal x[n].

A.   DTFT Formula:

ω = nω0 and X(ω)=Tcn

B.   Z Transform Formula List

1. Bilateral Z Transform Formula

$X(z)=Z\left \{ x[n] \right \}=\sum_{n=-\infty}^{\infty }x[n]z^{-n}$

2. Unilateral Z Transform Formula

$X(z)=Z\left \{ x[n] \right \}=\sum_{n=0}^{\infty }x[n]z^{-n}$

### Solved Z Transform Examples On How to Find Z Transform

Example: Express the z-transform for a finite series as given below.

x = {-2, -1, 1, 2, 3, 4, 5, 6, 7, 8}

Solution:

Known sequence of sample numbers x[n] = is x = {-2, -1, 1, 2, 3, 4, 5, 6, 7, 8}

Z-transform of x[n] can be mathematically expressed as:

X (z) = -2z0 – z-1 + z-2 + 2z-3 + 3z-4 + 4z-5 + 5z-6+ 6z-7+ 7z-8+ 8z-9

This can be further simplified as below.

X (z) = -2 – z-1 + z-2 + 2z-3 + 3z-4 + 4z-5 + 5z-6+ 6z-7+ 7z-8+ 8z-9

## FAQs on Z Transform Formula

1. What is Meant By Region of Convergence for a Z-transform?

Answer: As mentioned earlier, the z-transform comes in two parts. One is the region definition and other is a mathematical formula which is referred to as the region of convergence. This region of convergence requires to be described for all z-transforms, for values of z as it will describe the points on the z-plane where the z-transform converges and simply to say the z-transform exists.

Generally for systems we consist of transfer functions that have the forms:

H(z) = N(z)/D(z)

2. What are the Uses of Z Transformation?

Answer: The z-transform is quite a useful and a significant technique, applicable in areas of system designing, signal processing, and evaluation and control theory.

Moreover, frequency and the Time-domain are considered in the study of discrete-time signals and systems. The z-transform provides us a third representation for the study. But, all the three domains are linked with each other. A unique feature of the z-transform is that in reference to the signals and system of interest to us, all of the evaluations will be in terms of ratios of polynomials.


text:
An Empirical Determination of the Intergalactic Background Light from UV to FIR Wavelengths Using FIR Deep Galaxy Surveys and the Gamma-ray Opacity of the Universe

# An Empirical Determination of the Intergalactic Background Light from UV to FIR Wavelengths Using FIR Deep Galaxy Surveys and the Gamma-ray Opacity of the Universe

## Abstract

We have previously calculated the intergalactic background light (IBL) as a function of redshift from the Lyman limit in the far ultraviolet to a wavelength of 5 m near infrared range, based purely on data from deep galaxy surveys. Here we utilize similar methods to determine the mid- and far-infrared IBL from 5 m to 850 m. Our approach enables us to constrain the range of photon densities, by determining the uncertainties in observationally determined luminosity densities and spectral gradients. By also including the effect of the 2.7 K cosmic background photons, we determine upper and lower limits on the opacity of the universe to -rays up to PeV energies within a 68% confidence band.

Our direct results on the IBL are consistent with those from complimentary -ray analyses using observations from the Fermi -ray space telescope and the H.E.S.S. air Čerenkov telescope. Thus, we find no evidence of previously suggested processes for the modification of -ray spectra other than that of absorption by pair production alone.

diffuse radiation – galaxies: observations – gamma-rays: theory

## 1 Introduction

We have previously employed a fully empirical approach to calculating the intergalactic background light (IBL) to wavelengths up to 5 m by using observational data from deep galaxy surveys (Stecker, Malkan & Scully 2012 (hereafter SMS12); and Scully, Malkan & Stecker 2014 (hereafter SMS14)); for a similar approach, see also Helgason & Kashlinsky (2012) and Khaire & Srianand (2015). Here we extend our previous results from SMS12 and SMS14, determining both the intergalactic background light (IBL) at longer wavelengths out to 850 m and the subsequent -ray opacity of the Universe out to multi-TeV energies. The spectra of astrophysical sources of such high energy -rays are being studied by ground-based air Čerenkov telescopes and linked air Čerenkov telescope arrays.

We accomplish the goals of this paper by using very recent deep galaxy survey data at far infrared wavelengths where galaxy emission is produced by dust re-radiation rather than starlight. We also include the -ray opacity from photons of the 2.7 K cosmic background radiation (CBR). This enables us to extend our calculations of the -ray opacity of the Universe to energies much greater than a TeV.

Observations at wavelengths greater than 24 m, have been covered by the Multiband Imaging Photometer on the Spitzer space telescope (MIPS), now being dramatically advanced by the availability of data from the Photoconductor Array Camera and Spectrometer (PACS) and Spectral Photometric Imaging Receiver (SPIRE) instruments on the Herschel space telescope, the Planck space telescope, and by ground-based observations from the Atacama Large Millimeter Array (ALMA) and observations from the Balloon-borne Large Aperture Submillimeter Telescope (BLAST). By using only empirical data rather that modeling, our approach is, by definition, model independent. We use published luminosity functions and interpolations of luminosity densities between observed wavebands, observations from high-redshift galaxy surveys being now sufficient for this purpose.

Most past approaches to determining the IBL and its present-day spectral energy distribution, referred to as the ”extralgalactic background light (EBL)”, require assumptions about how the galaxy luminosity functions evolve (e.g. Malkan & Stecker 1998, 2001; Kneiske, Mannheim & Hartmann, 2002: Stecker, Malkan & Scully 2006 (SMS06); Franceschini et al. 2008; Finke, Razzaque, & Dermer, 2010; Kneiske & Dole 2010). Other approaches make use of semi-analytic models that require various assumptions regarding galaxy evolution, stellar population synthesis modeling, or star formation rates the properties of and dust attenuation, particularly for redshifts greater than 1 (e.g., Salamon & Stecker 1998; Gilmore et al. 2009; Somerville et al. 2012; Dominguez et al. 2011; Inoue et al. 2013; Driver et al. 2016). In contrast, our observationally-based approach is superior to model-based methods, since it enables a determination of both the IBL and its observational uncertainties. This is because we use observationally determined errors. Approaches that rely on modeling cannot determine such uncertainties.

Observational studies of blazar -ray spectra have also been used to probe the IBL (Ackermann et al. 2012; Abramowski et al. 2013; Biteau & Williams 2015) through its opacity effect caused by electron-positron pair production. This approach was originally suggested by Stecker, De Jager & Salamon (1992) when the infrared EBL was unknown. Our present method of determining the expected -ray opacity of the Universe by first using observational data to determine the IBL and then calculating its effect on -ray spectra of extragalactic sources complements the technique of using direct -ray observations. This is because the later approach requires a thorough knowledge of the intrinsic (unabsorbed) emission spectra of the -ray sources. This requirement introduces unavoidable uncertainties.

Our final results give the -ray opacity as a function of energy and redshift to within a 68% confidence band that is based on observational data. This opacity is solely caused by pair-production interactions of IBL photons with extragalactic -rays . Thus, a direct comparison of this effect with -ray spectra of extragalactic sources enables an assessment of possible additional spectral modifications. One such possible spectral modification has been suggested to be caused by secondary -ray production from cosmic-ray interactions along the line of sight to the source (Essey & Kusenko 2014). Another such modification might be caused by photon-axion oscillations during propagation from the source to the Earth (e.g., De Angelis et al. 2007; Mayer & Horns 2013).

In Section 2 we give the details of our calculations of upper and lower limits on IR galaxy luminosity densities for wavelengths between 5 m and 850 m as a function of redshift, paying particular attention to the effects of the emission spectra of polycyclic aromatic hydrocarbons (PAHs). In Section 3 we calculate the EBL based on the results from Section 2 and compare it with both other work and observational limits. In Section 4 we compute the resulting opacity of the universe to -rays out to a redshift of z = 5, including the total opacity from interactions with both IBL and thermal cosmic background radiation (CBR) photons. In our conclusion section, Section 5, we compare our results with those obtained by other methods in some of the papers mentioned above. We also discuss the implications of our results.

## 2 Calculating the Infrared IBL

We have previously calculated the intergalactic background light (IBL) as a function of redshift in the far ultraviolet to near infrared range, based purely on data from deep galaxy surveys (SMS12; SMS14). Here we utilize similar methods to extend the calculation into the mid- and far infrared IBL out to a wavelength of 850 m.

### 2.1 Determining the Luminosity Densities from Empirical Luminosity Functions

In our previous work the observationally tolerated ranges of photon densities were determined from the luminosity densities (LDs), , themselves, with errors provided by the various authors at wavelengths ranging from the far UV to the 5 m wavelength in the near IR. The LDs are computed by integrating fits to the observationally determined luminosity functions:

 ρLν=∫LmaxLmindLνLνΦ(Lν;z) (1)

Cases where authors did not directly compute the LDs were excluded because properly estimating their error requires knowledge of the covariance of the errors in the fit parameters of the luminosity functions (LFs) and also knowledge of any observational biases. To provide comprehensive redshift coverage of the LDs, SMS12 and SMS14 made use of continuum colors between the wavelength bands to fill in any gaps.

At wavelengths greater than 5 m very few studies provide determinations of luminosity densities (LDs) directly, as most authors are more concerned with calculating the total IR density, integrated over all wavelengths. This is because the total IR density is an observable that is correlated with the star formation rate. Therefore, in this paper we used observer-given analytic fits to the LFs at various wavelengths. When those fits were not provided, we ourselves fit the observed infrared galaxy LFs at various wavelengths and redshifts and then use equation (1) to obtain the LDs.

Galaxy LFs typically have characteristic shapes that are flatter at lower luminosities but fall off more steeply at the highest luminosities. However, at wavelengths greater than 5 m the Schechter function, commonly used at optical wavelengths, does not provide a good fit to the observed LFs, as its exponential decrease falls off too quickly at the bright end compared with the measured LFs. Most observers instead fit their LFs to a broken power-law function that describes the data better (e.g., Marchetti et al. 2016). The transition region between the power law that holds at low luminosities and that holding at high luminosities is usually referred to as the knee of the LF. The luminosity at the knee is usually designated as . Most of the luminosity density from an LF is produced by galaxies with luminosities in the vicinity of the knee; much fainter galaxies do not contribute much to the LD and the much brighter ones are quite rare. Thus, it is critical to determine the location of the knee in order to estimate the LDs with any accuracy.

For the cases where we were required to determine our own fits, we chose a double power-law fitting function of the form:

 Φ(L)=cL∗((LL∗)a+(LL∗)b) (2)

Here, the parameters and are the indices of the power-law fits in the low luminosity and high luminosity ranges respectively. The overall normalization of the LF is given by the parameter , which has the dimension of luminosity per Mpcdex. To compute the LDs, we integrate equation (1) over the galaxy luminosities between 4 10 L and 10 L.

In order to accurately compute the errors in the LDs as derived from the fit parameters, which is essential for our calculation, we must do so in a way that includes terms involving the off-diagonal elements of the error matrix of the fits so as to account for covariance. In cases where the authors provided there own fits, we have therefore re-derived the fits of the LFs to generate the error matrix, retaining the same choice of fitting function and fixed parameters, if given.

Our goal, as in SMS12 and SSM14, was to compute the observationally tolerated ranges of luminosity densities. This requires that we represent the error on these quantities as best we can. The statistical error is determined by properly propagating the fit parameter errors accounting for covariance, lest we overestimate the error. To compute the total error on each LD, we further added in quadrature an additional systematic error that accounts for cosmic variance. We compute cosmic variance based on the field sizes of the individual studies. For the AKARI Wide Field IR Survey Explorer (WISE) and GOODS fields this value is typically of order of 10%. In determining the LDs, for consistency, all observational LFs are scaled to a Hubble parameter of

Our new additions to our observationally determined LDs extend our coverage of rest frame galaxy photon production from the near IR to 850 m in the far IR, with enough determinations at each wavelength band to span the redshift range . Using published results derived from observations by the Spitzer and Herschel space telescopes, sufficient redshift coverage was found for wavelength bands of and m .

There were two cases where we were required to combine LFs from different observational studies in order to provide enough coverage to discriminate the location of particularly for redshifts greater than . At 12 m we combined the results of Perez-Gonzalez et al. (2005) and Rodighiero et al. (2010) to compute LDs in redshift bins centered on redshifts of and . In order to achieve sufficient redshift coverage at m we combined some of the higher redshift m data from Lapi et al. (2011) with that of the m data from Gruppioni et al. (2013). This gains us additional coverage at redshifts of and .

At 160, 350, 500, and 850 m , LF data only exists for the very nearby redshifts. We therefore assumed that their redshift evolution closely follows that of the 250 m band (Lapi et al. 2011; Marchetti et al. 2016). We use local LDs calculated in those bands as a normalization to this evolution. This assumption is justified because the emission in this wavelength region is dominated by warm dust. At 160 m we used the local LF given be Patel et al. (2013). At 350 and 500 m , we used the combined local LF data of Marchetti et al. (2016) from Herschel/SPIRE, the Herschel/SPIRE estimate of Vaccari et al. (2010), and the Planck satellite data from Negrello et al. (2013). At 850 m we computed the local LD from the LF provided by Negrello et al. (2013). Figures 1 and 2 show the resulting derived values for and their errors together with the observationally determined 1 confidence bands for the 8, 12, 15, 24 m bands and those of 35, 60, 90, and 250 m respectively.

Our calculations of LDs at mid-IR to far-IR wavelengths presented here is an extension of the work done in SMS12 and SSM14. Thus, the results given in those papers for wavelengths less than 5 m is almost unchanged from the results presented here. However, we have updated the far UV calculations to include the more recent work from Bouwens et al. (2015) for LDs in the redshift range of to . Even though the shape of the far UV band can affect other bands when filling in for redshift gaps using colors, the overall calculation yields results that are qualitatively the same as those presented in SMS12 and SSM14, as the newer data do not significantly change the general trend.

In order to place 68% upper and lower limits by using observational data on we make as few assumptions about the luminosity density evolution as possible. In SMS12 and SMS14 we utilized a robust rational fitting function in the form of a broken power-law dependent on in order to generate the confidence bands. We take the 68% confidence ranges of these fits in each waveband as the 1 confidence bands for the LDs. At redshifts beyond the redshift at the peak of star formation, the LDs decline with redshift. In accord with recent studies of the evolution of rest frame LFs in the UV that trace the star formation rate (Finkelstein et al. 2015), we conservatively assume upper and lower limit power-law functions in redshift to represent the rate of this decline as the highest redshifts. In the upper limit case, we assume a decline proportional to . In the lower limit case, we adopt a steeper decline proportional to . These assumptions have almost no impact on the derivation of the opacity confidence bands that we determined. Figures 1 and 2 show our results for LDs as a function of redshift at various wavelengths.

### 2.2 Taking account of PAH emission

At wavelengths greater than 20 m the LDs between the bands can be determined by smoothly interpolating between our observationally based LDs at specific wavelengths, since the spectral energy distributions (SEDs) from galaxies in this wavelength range are smooth modified blackbody spectra produced by dust re-radiation. However, in the 5 – 20 m range the situation is more complex.

In star-forming galaxies, the spectra between 7 and 13 m are dominated primarily by PAH emission. These PAH molecules are found in very small dust grains in intergalactic media. They absorb the UV photons emitted by hot young and stars and reemit them in molecular emission bands in the mid-IR. They are thus a strong signature of active star formation in galaxies (Peeters, Spoon & Tielens 2004). In this regard, we note that luminous star forming galaxies at higher redshift have more prominent PAH emission features. The importance of PAH features in the mid-IR at redshifts has been shown by Lagache et al. (2005).

The average SED of nearby star-forming galaxies (from Spoon et al. 2007, see also Smith et al. 2007) is shown in Figure 3 normalized to our best fit low redshift LD confidence band. One can see that there is a relative ”valley” between 9 and 11 m. A simple direct interpolation between our 8 and 12 m bands would therefore obtain an incorrectly high value for the LD in this wavelength range. Since we do not have wavelength coverage in this regime, we take this feature into account by lowering our interpolated LDs by a factor of 3 for the upper limit and a factor of 5 for the lower limit at 10 m . The factor of 5 is chosen as a lower limit based on the difference between the peak at 8 m and the depth of the valley near 10 m from the SED. For the upper limit, we relax the depth of this feature because, while star forming galaxies make up the bulk of the IBL, contributions from other galaxy types have a less pronounced PAH feature. Since, at high redshifts, PAH emission correlates with the star formation rate (Shipley et al. 2016), we assume that our relative PAH shape factors of 3 and 5 coevolve with redshift.

## 3 The Extragalactic Background Light

At wavelengths above 100 m the FIRAS and DIRBE instruments aboard the Cosmic Background Explorer (COBE) have measured the EBL (Fixsen et al. 1998; Lagache et al. 1999). This diffuse background has now been in large part resolved by recent galaxy count studies using the Herschel telescope (Berta et al. 2010; Béthermin et al. 2012; Viero et al. 2015) along with ground based studies using ALMA (Fujimoto et al. 2016) and BLAST (Marsden et al. 2011). These more recent studies strongly support the COBE results.

However, there are no direct measurements of the EBL in the infrared range below 100 m owing to the predominance of the foreground radiation of Zodiacal light from interplanetary dust re-radiation. The flux of Zodiacal light is approximately two orders-of-magnitude larger than the EBL flux (Spiesman et al. 1995). In this region only lower limits obtained from galaxy counts exist.

Using our calculations of the LDs in the mid-IR and far-IR as a function of wavelength, we have constructed a 68% confidence band for the spectral energy distribution of the cosmic diffuse infrared background light (the IBL at ). Figure 4 shows our new results, combined with our previous results at shorter wavelengths, taken from SMS12 and SMS14. The light shaded band shows the maximum effect of PAH emission. It can be seen that taking account of the details of the PAH spectrum does not significantly affect our EBL results. Figure 4 also shows the observational lower limits on the EBL obtained from galaxy counts (in blue), extrapolations of mid-IR galaxy counts from Spitzer, and direct measurements (in black).

## 4 The Optical Depth from γ+γ→e++e− Interactions with IBL and 2.7 K CBR Photons

The co-moving radiation energy density for wavelength at redshift , , where , is the time integral of the co-moving luminosity density ,

 uν(z)=∫zmaxzdz′ρν′(z′)dtdz(z′), (3)

where and is the redshift corresponding to initial galaxy formation (Salamon & Stecker 1998), and

 dtdz(z)=[H0(1+z)√ΩΛ+Ωm(1+z)3]−1, (4)

with and .

The upper and lower limits on our co-moving energy densities, derived using equation (3), are shown in Figures 5 and 6.

In calculating the -ray opacities we use the relations for the photon energy, , and the photon density, .

The cross section for photon-photon annihilation to electron-positron pairs was first calculated by Breit & Wheeler (1934) as a solid result of quantum electrodynamics. The threshold for this interaction follows from the Lorentz invariance of the square of the four-momentum vector. This reduces to the square of the threshold energy in the c.m.s., , that is necessary to produce twice the electron rest mass:

 s=2ϵEγ(1−cosθ)=4m2e (5)

The Lorentz invariant quantity given by has been determined to hold to better than one part in (Stecker & Glashow 2001; Jacobson, Liberati, Mattingly & Stecker 2004).

The optical depth for -rays caused by electron-positron pair production interactions with photons of the stellar radiation background is given by

 τ(E0,ze)=c∫ze0dzdtdz∫20dxx2∫∞0dν(1+z)3[uν(z)hν]σγγ[s=2E0hc/λx(1+z)], (6)

where is the co-moving energy density of the photon field, (Stecker, De Jager, & Salamon  1992).

In equations (5) and (6), is the observed -ray energy at redshift zero, is the wavelength at redshift , is the redshift of the -ray source at emission, ,
being the angle between the -ray and the soft background photon, and the pair production cross section is zero for center-of-mass energy , being the electron mass. Above this threshold, the pair production cross section is given by

 σγγ(s)=316σT(1−β2)[2β(β2−2)+(3−β4)ln(1+β1−β)], (7)

where is the Thompson scattering cross section and (Jauch & Rohrlich 1955).

It follows from equation (5) that the pair-production cross section has a threshold at .

The optical depth of the universe to the CBR is given by

 τCBR=5.00×105√1.11PeVEγ∫z0dz′ (1+z′) e−(1.11PeVEγ(1+z′)2)√ΩΛ+Ωm(1+z′)3 (8)

(Stecker 1969; SMS06).

Figure 7 shows the 68% confidence opacity bands for interactions with IBL photons given for sources at and , calculated using the methods described above, along with the opacity produced by interactions of -rays with photons of the 2.7 K cosmic background radiation. Note that at the higher energies and redshifts where the opacity is dominated by interactions with CRB photons, the uncertainty band becomes a very thin line, since the CBR-dominated opacity is exactly determined by equation (8).

## 5 Discussion and Conclusions

### 5.1 Comparison with our previous backward evolution model

Ten years ago we made estimates the diffuse infrared background when hardly any mid-IR and far-IR luminosity functions had been observationally determined at high redshifts (SMS06). That work was therefore based on the assumptions of a “backward evolution” model. Starting from the well-determined local (z=0) LF at 60m, we assumed that the average locally determined transformations between different mid-IR and far-IR wavelengths applied, unchanged, at all redshifts. The luminosity function used was a double power-law, similar to that of equation (2) with the parameters and W/Hz at as determined at 60 m, which was the wavelength for which the most complete galaxy LF existed. We then assumed that the effect of redshift evolution could be taken account purely by the evolution of , as described in SMS06.

We have compared the predictions of the SMS06 backwards evolution model with the recently observed IR LFs as used in this paper. If we make relatively small improvements in the assumed LF parameters, the backwards evolution model agrees with the new observations surprisingly well. The data favor a slightly flatter LF, with a low-luminosity slope of and a high-luminosity slope of , provided that we compensate by slightly decreasing the normalization of the LF at and to Mpc dex. We then match the observed LFs at higher redshifts, by assuming that increases as for , with constant at higher redshifts. In this way, we can obtain a good fit between the backward evolution model and the observational data.

Although this slightly modified backwards evolution reproduces the observed LFs at all redshifts well, there are a few discrepancies, in either direction. The most substantial disagreement is with the 8m observations of Huang et al. (2007) at . Below the knee of the LF the observed LF the data are up to 0.3 dex higher than the LF of our backwards evolution model. On the other hand, our best-fitting model overpredicts the 15 m LFs by up to 0.25 dex at luminosities above the knee as compared with the data of Pozzi et al. (2004), Le Floch et al (2005), and Rodighiero et al. (2010). The likely explanation of both of these discrepancies is that our previously proposed simple model SEDs are based on average 12m luminosities as derived from the 60 m observations. Therefore, they do not include the strong contributions from the PAH bands (See section 2.2). Thus our SEDs will overpredict the dust continuum on either side of the PAH bump feature as shown in Figure 3. Correspondingly, if we lower the normalization to better fit the shorter and longer wavelengths, we under-estimate the broadband fluxes at 8m rest wavelength.

Of course, our new observationally-based calculation presented in this paper avoids the problems of backward evolution models, since here we use directly observed LFs at 8, 15, and 24 m. These data include the effect of the PAH emission features and show their significance.

### 5.2 Our present results and comparison with other work

Figure 8 shows our 68% confidence band computed for the IBL (EBL) along with EBL SEDs obtained from the models of Franceschini et al. (2008) and Domínguez et al. (2011). Figure 9 shows our 68% opacity bands for and , in comparison with the opacity curves of Franceschini et al. (2008) and Domínguez et al. (2011). We note that Franceschini et al. (2008) used data only up to 8 m that were available at the time and did not include a PAH component in their model. The model of Domínguez et al. (2011) assumes a redshift evolution at redshifts greater than 1 that follows the evolution in the -band given by Cirasuolo et al. (2010).

### 5.3 Conclusion

In our previous papers (SMS12 and SMS14), we presented observationally based results for the IBL as a function of wavelength and redshift for wavelengths below of 5 m . Based on those results, we computed the -ray opacity of the Universe caused by electron-positron pair production up to a -ray energy of TeV. In this paper we have extended our determinations of the IBL within 68% confidence bands. This determination defines upper and lower limits on the IBL to longer wavelengths that extend to 850 m. Our model-independent results are based on observationally derived luminosity functions from recent galaxy survey data from both local and high redshift surveys. These data include results from Spitzer, Herschel and Planck. We then use these results to calculate the opacity of the Universe to -rays out to PeV energies. In doing so, we also take account of the redshift dependence of interactions of -rays with photons of the 2.7 K cosmic background radiation (CBR) (Stecker 1969), since the opacity from interactions with CBR photons dominates over that from interactions with IBL photons at the higher -ray energies and redshifts.

In figure 10 we give an energy-redshift plot showing the highest energy photons from extragalactic sources as a function of redshifts as determined from Fermi data (Abdo et al. 2010) plotted with our 68% confidence band for .

Our direct results on the IBL are consistent with those from complimentary -ray analyses using observations from the Fermi-LAT -ray space telescope and the H.E.S.S. air Čerenkov telescope. Figure 11 indicates how well our opacity results for overlap with those obtained by the Fermi collaboration (Ackermann et al. 2012). Our results are also compatible with those obtained from higher energy -ray observations using H.E.S.S. (Abramowski et al. 2013). This overlap of results from two completely different methods strengthens confidence that both techniques are indeed complimentary and supports the concept that the spectra of cosmic -ray sources can be used to probe the IBL (Stecker et al. 1992).

Thus, we find no evidence for modifications of -ray spectra by processes other than absorption by pair production, either by cosmic-ray interactions along the line of sight to the source (Essey & Kusenko 2014) or line-of-sight photon-axion oscillations during propagation (e.g., De Angelis et al. 2007; Mayer & Horns 2013). In this regard, we note that the Fermi Collaboration has very recently searched for irregularities in the -ray spectrum of NGC 1275 that would be caused by photon-axion oscillations and reported negative results (Ajello et al. 2016).

We conclude that modification of the high energy -ray spectra of extragalactic sources occurs dominantly by pair production interactions of these -rays with photons of the IBL. They therefore support the concept of using the future Čerenkov Telescope Array instruments to probe the cosmic background radiation fields at infrared wavelengths. This method can be used in conjunction with future deep galaxy survey observations using the near infrared and mid-infrared instruments aboard the James Webb Space Telescope.

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text:
Tutorial: Quick Start

This tutorial will address the fundamental steps when using 3DEC for the first time. It will help new users become comfortable with beginning new projects and developing a simple block model. A walk-through guide of a simple wedge failure example will be completed. Interactive controls using both the mouse and script commands input in the 3DEC console will be explained and illustrated.

Topics:

• rigid blocks
• joints
• histories
• GUI and plot manipulation

Commands: BLOCK CREATE, BLOCK DELETE, BLOCK FIX, BLOCK GROUP, BLOCK HIDE, BLOCK CUT, BLOCK PROPERTY, MODEL CYCLE, MODEL GRAVITY, MODEL SAVE

FISH: none

Files: “Users_Guide/getting_started/simple_wedge/simple_wedge.prj”
“Users_Guide/getting_started/simple_wedge/simple_wedge.dat”

Project

After starting 3DEC, you will be shown a dialog box with information about the version you are using. At the bottom of the dialog are several startup options related to projects. A project is simply a way to bundle together data files, FISH files, plots, etc. related to a problem being solved with 3DEC. Select “New Project …” from the prompt.

A save prompt will now appear. Click on the My Projects directory on the left, and save the new project as “getting_started.prj”. When you install 3DEC, a directory called My Projects is created for you in Documents/Itasca/3dec700. This is a useful place to store your own projects.

The commands used to complete this tutorial will be saved in a data (٭.dat) file. To create a new data file, select File → New Data File… or press Ctrl+N. Save the new data file as “simple_wedge.dat” in the same folder as the project file (“getting_started.prj”).

Layout

The 3DEC GUI (graphical user interface) contains display panels for data files, plots, the command console and project files. The layout of the panes in the 3DEC GUI can be customized to suit any user’s preference. There are several predefined layout options found under the “Layout” menu, including “Horizontal,” “Vertical,” “Single,” “Wide” and “Project” options. A 3DEC layout can also be saved and restored by setting up the preferred layout using the mouse and dragging different panes into place. Once the panes are in place, click “Layout” and select “Save layout….” A prompt will appear and request a save location for the layout file. This location is necessary when restoring saved layout preferences.

Choose Layout → Wide and your screen should appear as shown.

The newly created data file is listed under “Data Files,” on the left. The blank data file itself fills the main pane. The command console is at the bottom. When the data file is shown, a table of global FISH symbols is whon in the right. If a plot is shown, plot controls are on the right (select the Plot01 tab to see this).

Creating a Model

Now we can begin inputting commands to create the model. Commands can either by entered into the file and then executed later, or they can be entered one by one in the console at the bottom. It is recommended that commands be entered into the file because this will save them and allow you to easily run the same set of commands multiple times.

The following commands will clear previous model information, create the initial block and specify the block size. We specify a single polyhedral block using the block create brick command.

model new
block create brick (0,80) (-30,80) (0,50)

This will create a block with the dimensions 80 × 110 × 50. At any time, you can press Ctrl + spacebar to see keywords following a command. The keywords shown can be clicked and new keywords will then appear. In this way a command can be built up with minimal typing or memorizing of keywords. At any time it is also possible to press F1 to bring up the relevant section of the manual related to the command or keyword.

Execute the file by clicking on the green execute button.

To view the plot, select the Plot01 tab below the editor window. The plot controls will now appear on the right side. Select the Build Plot button as shown.

You will now see a ist of all possible plot items. Select Block and click OK.

The single block should now appear in the plot window. The default view of the model is oriented parallel to the $$xz$$-plane. Try holding down the right mouse button and moving the mouse, to rotate the plot. The figure below illustrates a rotated view of the block.

You can rename the plot by right-clicking on the Plot01 tab and selecting Rename. Change the plot name to Blocks.

Manipulating the view: Press and hold the right mouse button and drag the mouse across the plot to rotate the model; press and hold right-click + left-click to pan; zoom in and out using the scroll wheel. Press Ctrl + R to reset the view back to the xz plane. For more detailed instructions on view manipulation, right-click on the plot and select Plot Control Help.

Creating the Joint Sets

Joints are introduced to the model by using the block cut joint-set command. Joints can represent actual rock joints, or they can be used as “construction” cuts to create specific geometries. For this example, a series of joint sets are implemented in order to create the slope and wedge. We begin by cutting the joint horizontally in two locations using the keywords dip, dip-direction and origin. CAUTION: If no keywords are used, these properties will default to values of 0.

Click on the “simple_wedge” tab to go back to the data file and add the following commands.

block cut joint-set dip 90 dip-direction 180 origin 0,0,0
block cut joint-set dip 90 dip-direction 180 origin 0,50,0

The dip direction is degrees clockwise from North ($$y$$). The block is cut into three (3) sections along the $$y$$-axis. The middle block is assigned to a group named “inner block” using a RANGE command to specify a range in $$y$$-coordinates for the block centroids. The other blocks are called “outer blocks” to define the blocks that act as boundaries.

block group 'inner block' range pos-y 0 50
block group 'outer blocks' range group 'inner block' not

Execute the file. Click on the “Blocks” tab to see the block plot. You will now see three blocks of different colors. In the control panel on the right, click on the “Block Block” item. In “Attributes,” under “Label”, select Block Group.

The block plot will now look like this:

Additional joint sets are introduced into the model and create the sliding planes for the future wedge in both the horizontal and vertical directions. First, hide the outer blocks. A block hide command is used to do this. Hidden blocks are not cut by $$block cut$$ commands. Two (2) shallow-dipping fracture planes are created, and then five (5) equally spaced high angle foliation planes are made.

block hide range group 'outer blocks'
block cut joint-set dip 2.5 dip-direction 235 origin 30,0,12.5
block cut joint-set dip 2.5 dip-direction 315 origin 35,0,30
block cut joint-set dip 76 dip-direction 270 spacing 4 num 5 origin 38,0,12.5

The use of the spacing keyword specifies an average spacing distance between joints. The num keyword defines the number of joints in the set. Execute the file and click on the “Blocks” plot tab. Change the “Label” back to “Block”. The model should look similar to the figure below.

We now hide the slope blocks and create a horizontal joint plane which will make up the base of the slope excavation.

block hide range pos-x 30,80 pos-y 0,50 pos-z 0,50
block cut joint-set dip 0 dip-direction 0 origin 0,0,10
block hide range pos-z 0,10
block group 'excavate'

We assign the blocks within the excavate region to a group called “excavate.”

Creating the Wedge Feature

Finally, we create the joint planes that define the wedge in the slope using the following commands:

block hide off
block hide range group 'outer blocks'
block hide range pos-z 0,10
block hide range pos-x 55,80
block hide range pos-x 0,30
block cut joint-set dip 70 dip-direction 200 origin 0,35,0
block cut joint-set dip 60 dip-direction 330 origin 50,15,50

Executing the file at this point will show the blocks with the high angle foliations and the planes that define the wedge as shown below.

Boundary Conditions

Next, the boundary blocks are immobilized. Using the block fix command and specifying the necessary range will freeze each block’s centroid velocity at its current value. In this case, the centroid velocity is currently zero. This will prevent any movement, thus creating boundary conditions. If you fix a block that is already moving, the block will continue to move at that current velocity. The base of the model and the blocks behind the foliations are fixed in place. The outer blocks are also fixed. The “excavate” group is deleted from the model. The outer blocks are left hidden for plotting purposes.

block hide off
block fix range pos-z 0 10
block fix range pos-x 55 80
block fix range group 'outer blocks'
block hide range group 'outer blocks'
block delete range group 'excavate'

Go back to the “Blocks” plot. Add axes to the plot by selecting the “Build Plot” button (+) and choosing Misc → Axes.

The axes can be resized by left-clicking and dragging the edges of the item. They can be moved by left-clicking and dragging the middle of the item. The plot should now look similar to the plot below.

Gravity

The model gravity command assigns a gravitational acceleration in the negative z-direction. In this example, we specify a value of 10 m/sec2. The $$x$$ and $$y$$ components are 0.

model gravity 0 0 -10

Material Properties

Note

٭Mass density of the block material, not the unit weight, is assigned.

The material properties are assigned to the blocks and joints. It is first necessary to show the hidden blocks so that they are assigned properties too.

In this example, the mass density٭ of all blocks is specified to be 2,000 units (kg/m3, in this case). Blocks are assumed to be rigid; block deformability is neglected.

block hide off
block property density 2000

Joint properties are assigned next. Before joint properties can be assigned, subcontacts need to be generated with the command block contact generate-subcontacts. This command triangulates the block surfaces and adds subcontacts at the triangle vertices. It is necessary to have multiple subcontacts per contact to resist rotation. Note that this step is not necessary if you are using deformable blocks and you have given the block zone generate command to create zones. In this case, the face triangulation matches the zone faces. All joints have the same normal and shear stiffness, equal to 1.0 × 10^9 units (Pa/m, in this case; see Joint Properties for a discussion on how to choose joint stiffness values). The true joints assigned a friction angle of 89º (for now), and the joints between the inner and outer blocks are assigned a friction angle of 0º. The “boundary” joints are identified with a group name so we can identify them again later.

block contact generate-subcontacts
block contact prop stiffness-norm=1e9 stiffness-shear=1e9 friction=89
block contact group 'boundary' range orientation dip 90 dip-dir 180
block contact prop fric=0.0 range group 'boundary'

If running the model in large strain, it is also necessary to define the properties for any new contacts that may form when blocks are sliding. This is done with the following command:

block contact material-table default prop stiffness-norm=1e9 stiffness-shear=1e9 fric=0

Note

It is possible in 3DEC to abbreviate the commands, as long as there are enough characters that they can be uniquely identified. For example, property can be abbreviated as prop.

It is possible to specify different contact properties for different blocks that come into contact, hence the need for a “material table”. In this case, we can assume that all new contacts are frictionless.

Note that different constitutive models are available for joints. In this example, a simple Mohr-Coulomb model is used (the default).

Solution

The problem is ready to be solved. It is often helpful to judge material behavior by observing the motion of specified points in the rock mass. In this problem, we monitor the $$z$$-velocity of a vertex nearest the point $$x$$ = 40, $$y$$ = 40, $$z$$ = 50. In this example, we need to be careful to select a point on the wedge and not the corresponding point in the host rock. For this reason we will first hide the blocks on one side of the joint as shown before taking the history.

block hide range plane dip 70 dip-direction 200 origin 0,35,0 below
block history vel-z position (40,40,50)

Following execution of this command, the program returns information about the selected monitoring point.

We then show the blocks again, hide the outer blocks:

block hide off
block hide range group 'outer blocks'

Finally, we bring the model to equilibrium with the following commands. Before cycling, we need to indicate if we are running in large or small strain using the model large-strain command. We can then use the model solve command to reach equilibrium. Note that the blocks won’t slide because we have initialy specified a high friction angle.

model large-strain on
model solve

Execute the file by clicking the green execute button at the top. During execution, the current cycle count, the calculation time, the maximum out-of-balance force (Unbalanced force is defined in Reaching Equilibrium), and the clock time are printed on the screen every 10 cycles. Inspection of these values indicates that equilibrium has been achieved (unbalanced force approaches zero).

Create a new plot called “Vel His” by selecting File → New Plot and entering the name at the prompt. In the control panel, select the “Build Plot” button (+) and then choose Charts → History Chart, and a blank history chart will appear. Next, under the “Attributes” tab, add the “Z-velocity” history by clicking the blue “+” symbol. Note that each history is assigned an ID when it is created. Since we only have one history, it is assigned an ID of 1. The blue “+” symbol allows you to add the history of ID 1 to the plot.

Change the label for the $$y$$-axis by selecting Y-Axis → Label and entering “vertical velocity.”

The plot should look like this:

Plot Image Files

After the model has been brought to equilibrium, creating an image file of a plot of the initial state may be useful. A plot title can be provided using the model title command and including a string afterwards:

model title 'Rock Slope Stability'

The title will be saved but not displayed to the plot. In order to display the title on the plot, click the “View Settings” button above the plot items (it looks like a check mark in a box). Then check the “Job Title” box in the resulting dialog.

The title will appear at the top of the plot, above the history. The properties of the title such as size, font, style (bold, italic, bold italic) and color can be modified in the Options dialog.

The plot can then be exported as several different file types. Navigate under File → Vel His → Export → Bitmap, or simply right-click on the plot and select Export → Bitmap. A plot and file name must be specified. It is also possible to export the history data as comma-separated values in a text file by right-clicking and selecting Export → CSV File.

Save Initial State

It is good practice to save the initial state of the model so that it can be restarted at any time (for example, to perform parameter studies). We save the current state to a file called “slope-initial.sav” by typing

model save 'slope-initial'

The file name extension .sav is automatically added if no extension is given.

Reduce Joint Friction and Solve

The behavior of the slope can be studied by reducing the friction angle of the joints. We reduce the friction angle to 6º using the following commands (either add to the file and re-execute, or simply type the command into the console at the bottom of the 3DEC GUI):

block contact prop fric=6.0 range group 'boundary' not

The next step is to continue the calculations with the reduced friction for an additional 2000 cycles. We don’t want to use model solve since it is unlikely that the model will reach equilibrium. The failure can be seen on the “Blocks” plot. For plotting purposes, turn on the job title by going to “View Settings” in the control panel (the check-mark in a box) and checking the “Job Title” option. This will add the job title specified previously. To add a title specific to this plot, check on the “Plot Title” option and enter some text (e.g., “Wedge Failure”).

To watch the failure progress, type the following command in the console at the bottom of the screen and press Enter:

model cycle 2000

Upon executing the problem, the block plot will illustrate the wedge failure caused by the reduction in the friction angle. The failure mode combines rotational failure along the foliation planes and rotational failure of the wedge. The wedge failure dominates the failure, as shown by the block plot. The rotational mechanism contributes to the collapse.

Cross Section

A vertical cross-section plot taken through the model will demonstrate the rotational mechanism. Click File → New Plot and name the plot “Cross Section”. From the list of plot items, select Blocks → Vectors. By default, the vectors will show displacement, but this can be changed with the Type attribute. Under the “Attributes”, toggle the “By-Magnitude” option on.

Add blocks to the plot. The fill is toggled off by clicking on the Polygons → Fill box in the “Attributes”.

Finally, a cutting plane is required to create the cross section. This is done by clicking on the “Cutting Tool” under the “Block Block” plot item. By default, the plane will cut perpendicular to the current view, half-way through the model. So in this case, the plane is normal to the $$y$$-axis (0,1,0) and the origin is at $$y$$ = 25.

Also turn on the cut plane for the Block Displacement Vectors plot item. The resulting cross section displaying the displacement vectors:

Conclusion

This concludes the getting started tutorial. Save the project by clicking File → Save Project. You will be asked if you also want to save the model state. Click “Yes” and save the model as ‘slope-final’. From this point forward, you may wish to try other various 3DEC features in an attempt to stabilize the slope. You can restart from the initial state by entering

model restore 'slope-initial'

There are structural elements that can be incorporated and are described in Structural Elements.


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## New answers tagged combustion

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Since the reaction is done in excess of air, you can assume complete combustion and accordingly you should break down the reaction to only this. Apparently the question demands an answer in the following form: $$\ce{C8H18 + 1.2a (O2 + 3.76 N2) -> b CO2 + c H2O + d O2 + e N2},$$ however, this is the same as $$\begin{multline} \ce{C8H18 + x O2 + 0.2x O2 + ... -2 Key things I did not consider properly: 120% of excess air, which means 0.2 of oxygen goes free, this changes my equation to:$$\ce{C8H18 + 1.2$a$ (O2 + 3.76 N2) -> b CO2 +c H2O + 0.2a O2 + e N2} So, \begin{align} \ce{C}:&& b &= 8\\ \ce{H}:&& 2c &= 18\\ \ce{O}:&& 2b + c + 2(0.2) &= 2\times1.2a \implies a = 12.5\\ \... 1 I'll continue after volume of \ce H_2O vapours = 0.5 ml Vapour pressure is equal to the partial pressure of water in gas phase. P_{\ce H_2O} = P_{total} × mole fraction of \ce H_2O Where P_{total} is 1 atm and mole fraction = 0.5/25 SoP_{\ce H_2O} = \frac{1}{50} atm = \frac{760}{50} mm Hg Now for the solution of non volatile solute $A$ ...

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# Definition of Infimum. Meaning of Infimum. Synonyms of Infimum

Here you will find one or more explanations in English for the word Infimum. Also in the bottom left of the page several parts of wikipedia pages related to the word Infimum and, of course, Infimum synonyms and on the right images related to the word Infimum.

## Definition of Infimum

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## Meaning of Infimum from wikipedia

- In mathematics, the infimum (abbreviated inf; plural infima) of a subset S of a partially ordered set T is the greatest element in T that is less than...
- fashion for a function (see limit of a function). For a set, they are the infimum and supremum of the set's limit points, respectively. In general, when...
- concepts of essential supremum and essential infimum are related to the notions of supremum and infimum, but adapted to measure theory and functional...
- unique supremum (also called a least upper bound or join) and a unique infimum (also called a greatest lower bound or meet). An example is given by the...
- if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum. If a sequence of real numbers is increasing and bounded...
- partially ordered set in which all subsets have both a supremum (join) and an infimum (meet). Complete lattices appear in many applications in mathematics and...
- (f\ominus b)(x)=\inf _{y\in B}[f(x+y)-b(y)]} , where "inf" denotes the infimum. In other words the erosion of a point is the minimum of the points in...
- Dirichlet's integral is bounded from below, the existence of an infimum is guaranteed. That this infimum is attained was taken for granted by Riemann (who coined...
- limit infimum and limit supremum of a set sequence always exist and can be used to determine convergence: the limit exists if the limit infimum and limit...
- are respectively the supremum (least upper bound) of S, denoted ⋁S, and infimum (greatest lower bound) of S, denoted ⋀S. In general, the join and meet...
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