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.