\displaystyle \text{Solve the following differential equations:}

\displaystyle \textbf{Question 1: }~x^{2}\,dy+y(x+y)\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x^{2}\,dy+y(x+y)\,dx=0
\displaystyle \Rightarrow x^{2}\,dy=-y(x+y)\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{y(x+y)}{x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=-\frac{vx(x+vx)}{x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=-v(1+v)
\displaystyle \Rightarrow x\frac{dv}{dx}=-v-v^{2}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-(v^{2}+2v)
\displaystyle \Rightarrow \frac{dv}{v(v+2)}=-\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{dv}{v(v+2)}=-\int \frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\int\left(\frac{1}{v}-\frac{1}{v+2}\right)dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\left(\log|v|-\log|v+2|\right)=-\log|x|+C
\displaystyle \Rightarrow \log\!\left|\frac{v}{v+2}\right|=2\log\!\left|\frac{C}{x}\right|
\displaystyle \Rightarrow \log\!\left|\frac{v}{v+2}\right|=\log\!\left|\frac{C^{2}}{x^{2}}\right|
\displaystyle \Rightarrow \frac{v}{v+2}=\frac{C^{2}}{x^{2}}
\displaystyle \Rightarrow \frac{y/x}{y/x+2}=\frac{C^{2}}{x^{2}}
\displaystyle \Rightarrow \frac{y}{y+2x}=\frac{C^{2}}{x^{2}}
\displaystyle \Rightarrow x^{2}y=C^{2}(y+2x)
\displaystyle \Rightarrow x^{2}y=K(y+2x)\quad (\text{where } K=C^{2})

\displaystyle \textbf{Question 2: }~\frac{dy}{dx}=\frac{y-x}{y+x}. \hspace{2.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{y-x}{y+x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx-x}{vx+x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{v+1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v-1}{v+1}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v-1-v(v+1)}{v+1}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{2}+1}{v+1}
\displaystyle \Rightarrow \frac{v+1}{v^{2}+1}\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{v+1}{v^{2}+1}\,dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \int \frac{v}{v^{2}+1}\,dv+\int \frac{1}{v^{2}+1}\,dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{2}\log(v^{2}+1)+\tan^{-1}v=-\log|x|+C
\displaystyle \Rightarrow \log(v^{2}+1)+2\tan^{-1}v=2\log|x|+C
\displaystyle \Rightarrow \log\!\big((v^{2}+1)x^{2}\big)+2\tan^{-1}v=C
\displaystyle \Rightarrow \log(y^{2}+x^{2})+2\tan^{-1}\!\left(\frac{y}{x}\right)=C

\displaystyle \textbf{Question 3: }~\frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-x^{2}}{2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v^{2}-1}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-1}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{2}+1}{2v}
\displaystyle \Rightarrow \frac{2v}{v^{2}+1}\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2v}{v^{2}+1}\,dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \log(v^{2}+1)=-\log|x|+C
\displaystyle \Rightarrow \log(v^{2}+1)=\log\!\left(\frac{C}{|x|}\right)
\displaystyle \Rightarrow v^{2}+1=\frac{C}{x}
\displaystyle \Rightarrow \left(\frac{y}{x}\right)^{2}+1=\frac{C}{x}
\displaystyle \Rightarrow y^{2}+x^{2}=Cx

\displaystyle \textbf{Question 4: }~x\frac{dy}{dx}=x+y.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\frac{dy}{dx}=x+y
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+y}{x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+vx}{x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=1+v
\displaystyle \Rightarrow x\frac{dv}{dx}=1
\displaystyle \Rightarrow dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow v=\log|x|+C
\displaystyle \Rightarrow \frac{y}{x}=\log|x|+C
\displaystyle \Rightarrow y=x\log|x|+Cx

\displaystyle \textbf{Question 5: }~(x^{2}-y^{2})\,dx-2xy\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{2}-y^{2})\,dx-2xy\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}-y^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}-v^{2}x^{2}}{2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1-v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v^{2}}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-3v^{2}}{2v}
\displaystyle \Rightarrow \frac{2v}{1-3v^{2}}\,dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2v}{1-3v^{2}}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{3}\log|1-3v^{2}|=\log|x|+C
\displaystyle \Rightarrow \log|1-3v^{2}|=-3\log|Cx|
\displaystyle \Rightarrow \log|1-3v^{2}|=\log\!\left|\frac{1}{(Cx)^{3}}\right|
\displaystyle \Rightarrow 1-3v^{2}=\frac{1}{(Cx)^{3}}
\displaystyle \Rightarrow 1-3\left(\frac{y}{x}\right)^{2}=\frac{1}{(Cx)^{3}}
\displaystyle \Rightarrow x(x^{2}-3y^{2})=\frac{1}{C^{3}}
\displaystyle \Rightarrow x(x^{2}-3y^{2})=K\quad (\text{where } K=\frac{1}{C^{3}})

\displaystyle \textbf{Question 6: }~\frac{dy}{dx}=\frac{x+y}{x-y}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{x+y}{x-y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v-v(1-v)}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{1-v}
\displaystyle \Rightarrow \frac{1-v}{1+v^{2}}\,dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1-v}{1+v^{2}}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow \int \frac{1}{1+v^{2}}\,dv-\int \frac{v}{1+v^{2}}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow \tan^{-1}v-\frac{1}{2}\log(1+v^{2})=\log|x|+C
\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{y}{x}\right)-\frac{1}{2}\log\!\left(1+\frac{y^{2}}{x^{2}}\right)=\log|x|+C
\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{y}{x}\right)-\frac{1}{2}\log\!\left(\frac{x^{2}+y^{2}}{x^{2}}\right)=\log|x|+C
\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{y}{x}\right)-\frac{1}{2}\log(x^{2}+y^{2})+\log|x|=\log|x|+C
\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{y}{x}\right)=\frac{1}{2}\log(x^{2}+y^{2})+C

\displaystyle \textbf{Question 7: }~2xy\frac{dy}{dx}=x^{2}+y^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle 2xy\frac{dy}{dx}=x^{2}+y^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^{2}}{2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v^{2}}{2v}
\displaystyle \Rightarrow \frac{2v}{1-v^{2}}\,dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2v}{1-v^{2}}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow -\log|1-v^{2}|=\log|x|+C
\displaystyle \Rightarrow -\log|1-v^{2}|-\log|x|=C
\displaystyle \Rightarrow \log|x(1-v^{2})|=C
\displaystyle \Rightarrow x(1-v^{2})=C
\displaystyle \Rightarrow x\left(1-\frac{y^{2}}{x^{2}}\right)=C
\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{x}=C
\displaystyle \Rightarrow x^{2}-y^{2}=Cx

\displaystyle \textbf{Question 8: }~x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}-2y^{2}+xy}{x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}-2v^{2}x^{2}+vx^{2}}{x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=1-2v^{2}+v
\displaystyle \Rightarrow x\frac{dv}{dx}=1-2v^{2}
\displaystyle \Rightarrow \frac{1}{1-2v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{1-2v^{2}}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \frac{1}{2\sqrt{2}}\log\!\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=\log|x|+C
\displaystyle \Rightarrow \log\!\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=2\sqrt{2}\log|x|+C
\displaystyle \Rightarrow \frac{1+\sqrt{2}v}{1-\sqrt{2}v}=(Cx)^{2\sqrt{2}}
\displaystyle \Rightarrow \frac{x+\sqrt{2}y}{x-\sqrt{2}y}=(Cx)^{2\sqrt{2}}

\displaystyle \textbf{Question 9: }~xy\frac{dy}{dx}=x^{2}-y^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle xy\frac{dy}{dx}=x^{2}-y^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}-y^{2}}{xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}-v^{2}x^{2}}{vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1-v^{2}}{v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-v^{2}}{v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-2v^{2}}{v}
\displaystyle \Rightarrow \frac{v}{1-2v^{2}}\,dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{v}{1-2v^{2}}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{4}\log|1-2v^{2}|=\log|x|+C
\displaystyle \Rightarrow \log|1-2v^{2}|=-4\log|x|+C
\displaystyle \Rightarrow \log\!\big((1-2v^{2})x^{4}\big)=C
\displaystyle \Rightarrow (1-2v^{2})x^{4}=C
\displaystyle \Rightarrow \left(1-2\frac{y^{2}}{x^{2}}\right)x^{4}=C
\displaystyle \Rightarrow x^{2}(x^{2}-2y^{2})=C

\displaystyle \textbf{Question 10: }~y e^{x/y}\,dx=(x e^{x/y}+y)\,dy.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle ye^{x/y}\,dx=(xe^{x/y}+y)\,dy
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{xe^{x/y}+y}{ye^{x/y}}
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{x}{y}+e^{-x/y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } x=vy \text{ and } \frac{dx}{dy}=v+y\frac{dv}{dy}, \text{ we get}
\displaystyle v+y\frac{dv}{dy}=v+e^{-v}
\displaystyle \Rightarrow y\frac{dv}{dy}=e^{-v}
\displaystyle \Rightarrow e^{v}\,dv=\frac{1}{y}\,dy
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int e^{v}\,dv=\int \frac{1}{y}\,dy
\displaystyle \Rightarrow e^{v}=\log|y|+C
\displaystyle \Rightarrow e^{x/y}=\log|y|+C

\displaystyle \textbf{Question 11: }~x^{2}\frac{dy}{dx}=x^{2}+xy+y^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x^{2}\frac{dy}{dx}=x^{2}+xy+y^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}+xy+y^{2}}{x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}+vx^{2}+v^{2}x^{2}}{x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=1+v+v^{2}
\displaystyle \Rightarrow x\frac{dv}{dx}=1+v^{2}
\displaystyle \Rightarrow \frac{1}{1+v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{1+v^{2}}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \tan^{-1}v=\log|x|+C
\displaystyle \Rightarrow \tan^{-1}\!\left(\frac{y}{x}\right)=\log|x|+C

\displaystyle \textbf{Question 12: }~(y^{2}-2xy)\,dx=(x^{2}-2xy)\,dy.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (y^{2}-2xy)\,dx=(x^{2}-2xy)\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}-2xy}{x^{2}-2xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-2vx^{2}}{x^{2}-2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v^{2}-2v}{1-2v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-2v}{1-2v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{3v^{2}-3v}{1-2v}
\displaystyle \Rightarrow \frac{1-2v}{3v^{2}-3v}\,dv=\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1-2v}{3v^{2}-3v}\,dv=\int \frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{3}\int \frac{6v-3}{3v^{2}-3v}\,dv=\int \frac{dx}{x}
\displaystyle \text{Let } 3v^{2}-3v=t
\displaystyle \Rightarrow (6v-3)\,dv=dt
\displaystyle \Rightarrow -\frac{1}{3}\int \frac{1}{t}\,dt=\int \frac{dx}{x}
\displaystyle \Rightarrow -\frac{1}{3}\log|t|=\log|x|+C
\displaystyle \Rightarrow -\frac{1}{3}\log|3v^{2}-3v|=\log|x|+C
\displaystyle \Rightarrow -\frac{1}{3}\log|v^{2}-v|=\log|x|+C
\displaystyle \Rightarrow \log|v^{2}-v|=-3\log|x|+C
\displaystyle \Rightarrow \log\!\left(\frac{v^{2}-v}{x^{3}}\right)=C
\displaystyle \Rightarrow \frac{v^{2}-v}{x^{3}}=C
\displaystyle \Rightarrow \frac{(y/x)^{2}-(y/x)}{x^{3}}=C
\displaystyle \Rightarrow y^{2}-xy=Cx^{3}
\displaystyle \Rightarrow x^{2}y-xy^{2}=K\quad (\text{where } K=C)

\displaystyle \textbf{Question 13: }~2y\,dx+(x^{2}+2y^{2})\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle 2xy\,dx+(x^{2}+2y^{2})\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{2xy}{x^{2}+2y^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=-\frac{2vx^{2}}{x^{2}+2v^{2}x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=-\frac{2v}{1+2v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{2v}{1+2v^{2}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{3v+2v^{3}}{1+2v^{2}}
\displaystyle \Rightarrow \frac{1+2v^{2}}{3v+2v^{3}}\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1+2v^{2}}{3v+2v^{3}}\,dv=-\int \frac{dx}{x}
\displaystyle \text{Putting } 3v+2v^{3}=t
\displaystyle \Rightarrow (3+6v^{2})\,dv=dt
\displaystyle \Rightarrow \frac{1}{3}\int \frac{dt}{t}=-\int \frac{dx}{x}
\displaystyle \Rightarrow \frac{1}{3}\log|t|=-\log|x|+C
\displaystyle \Rightarrow \log|3v+2v^{3}|=-3\log|x|+C
\displaystyle \Rightarrow \log\!\big((3v+2v^{3})x^{3}\big)=C
\displaystyle \Rightarrow (3v+2v^{3})x^{3}=C
\displaystyle \Rightarrow (3y/x+2y^{3}/x^{3})x^{3}=C
\displaystyle \Rightarrow 3yx^{2}+2y^{3}=C

\displaystyle \textbf{Question 14: }~3x^{2}\,dy=(3xy+y^{2})\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle 3x^{2}\,dy=(3xy+y^{2})\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3xy+y^{2}}{3x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{3vx^{2}+v^{2}x^{2}}{3x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{3v+v^{2}}{3}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}}{3}
\displaystyle \Rightarrow \frac{3}{v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{3}{v^{2}}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{3}{v}=\log|x|+C
\displaystyle \Rightarrow -\frac{3x}{y}=\log|x|+C

\displaystyle \textbf{Question 15: }~\frac{dy}{dx}=\frac{x}{2y+x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{x}{2y+x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x}{2vx+x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1}{2v+1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1}{2v+1}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-2v^{2}-v}{2v+1}
\displaystyle \Rightarrow \frac{2v+1}{1-2v^{2}-v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2v+1}{1-2v^{2}-v}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \int \frac{2v+1}{(2v-1)(v+1)}\,dv=-\int \frac{1}{x}\,dx
\displaystyle \text{Let } \frac{2v+1}{(2v-1)(v+1)}=\frac{A}{2v-1}+\frac{B}{v+1}
\displaystyle \Rightarrow 2v+1=A(v+1)+B(2v-1)
\displaystyle \Rightarrow A+2B=2,\; A-B=1
\displaystyle \Rightarrow A=\frac{4}{3},\; B=\frac{1}{3}
\displaystyle \Rightarrow \int \frac{2v+1}{(2v-1)(v+1)}\,dv=\frac{4}{3}\int \frac{1}{2v-1}\,dv+\frac{1}{3}\int \frac{1}{v+1}\,dv
\displaystyle \Rightarrow \frac{2}{3}\log|2v-1|+\frac{1}{3}\log|v+1|=-\log|x|+C
\displaystyle \Rightarrow \log\!\big((2v-1)^{2}(v+1)\big)=-3\log|x|+C
\displaystyle \Rightarrow (2v-1)^{2}(v+1)=\frac{C}{x^{3}}
\displaystyle \Rightarrow \left(2\frac{y}{x}-1\right)^{2}\left(\frac{y}{x}+1\right)=\frac{C}{x^{3}}
\displaystyle \Rightarrow (x+y)(2y-x)^{2}=C

\displaystyle \textbf{Question 16: }~(x+2y)\,dx-(2x-y)\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x+2y)\,dx-(2x-y)\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+2y}{2x-y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+2vx}{2x-vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+2v}{2-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{2-v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{2-v}
\displaystyle \Rightarrow \frac{2-v}{1+v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2-v}{1+v^{2}}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \int \frac{2}{1+v^{2}}\,dv-\int \frac{v}{1+v^{2}}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow 2\tan^{-1}v-\frac{1}{2}\log(1+v^{2})=\log|x|+C
\displaystyle \Rightarrow 2\tan^{-1}v=\log|x|+C+\frac{1}{2}\log(1+v^{2})
\displaystyle \Rightarrow 2\tan^{-1}v=\log\!\left(Cx\sqrt{1+v^{2}}\right)
\displaystyle \Rightarrow Cx\sqrt{1+v^{2}}=e^{2\tan^{-1}v}
\displaystyle \Rightarrow C\sqrt{x^{2}+y^{2}}=e^{2\tan^{-1}\!\left(\frac{y}{x}\right)}

\displaystyle \textbf{Question 17: }~\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{y}{x}\sqrt{\frac{y^{2}}{x^{2}}-1}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=v-\sqrt{v^{2}-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\sqrt{v^{2}-1}
\displaystyle \Rightarrow \frac{1}{\sqrt{v^{2}-1}}\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{\sqrt{v^{2}-1}}\,dv=-\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \log\!\left|v+\sqrt{v^{2}-1}\right|=-\log|x|+C
\displaystyle \Rightarrow \log\!\left|(v+\sqrt{v^{2}-1})x\right|=C
\displaystyle \Rightarrow (v+\sqrt{v^{2}-1})x=C
\displaystyle \Rightarrow \left(\frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}-1}\right)x=C
\displaystyle \Rightarrow y+\sqrt{y^{2}-x^{2}}=C

\displaystyle \textbf{Question 18: }~\frac{dy}{dx}=\frac{y}{x}\left(\log y-\log x+1\right).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{y}{x}\{\log\!\left(\frac{y}{x}\right)+1\}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=v(\log v+1)
\displaystyle \Rightarrow x\frac{dv}{dx}=v\log v+v-v
\displaystyle \Rightarrow x\frac{dv}{dx}=v\log v
\displaystyle \Rightarrow \frac{1}{v\log v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{v\log v}\,dv=\int \frac{1}{x}\,dx
\displaystyle \text{Putting } \log v=t
\displaystyle \Rightarrow dv=v\,dt
\displaystyle \Rightarrow \int \frac{1}{v t}\,v\,dt=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \int \frac{1}{t}\,dt=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \log|t|=\log|x|+C
\displaystyle \Rightarrow t=Cx
\displaystyle \Rightarrow \log v=Cx
\displaystyle \Rightarrow \log\!\left(\frac{y}{x}\right)=Cx

\displaystyle \textbf{Question 19: }~\frac{dy}{dx}=\frac{y}{x}+\sin\left(\frac{y}{x}\right).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{y}{x}+\sin\!\left(\frac{y}{x}\right)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=v+\sin v
\displaystyle \Rightarrow x\frac{dv}{dx}=\sin v
\displaystyle \Rightarrow \frac{1}{\sin v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{\sin v}\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \int \mathrm{cosec}\,v\,dv=\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \log\!\left|\tan\!\left(\frac{v}{2}\right)\right|=\log|x|+C
\displaystyle \Rightarrow \tan\!\left(\frac{v}{2}\right)=Cx
\displaystyle \Rightarrow \tan\!\left(\frac{y}{2x}\right)=Cx

\displaystyle \textbf{Question 20: }~y^{2}\,dx+(x^{2}-xy+y^{2})\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y^{2}\,dx+(x^{2}-xy+y^{2})\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{y^{2}}{x^{2}-xy+y^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=-\frac{v^{2}x^{2}}{x^{2}-vx^{2}+v^{2}x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=-\frac{v^{2}}{1-v+v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{2}}{1-v+v^{2}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v+v^{3}}{1-v+v^{2}}
\displaystyle \Rightarrow \frac{1-v+v^{2}}{v(1+v^{2})}\,dv=-\frac{dx}{x}
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1-v+v^{2}}{v(1+v^{2})}\,dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \int \frac{1}{v}\,dv-\int \frac{v}{1+v^{2}}\,dv=-\int \frac{dx}{x}
\displaystyle \Rightarrow \log|v|-\tan^{-1}v=-\log|x|+C
\displaystyle \Rightarrow \log|v|+\log|x|=\tan^{-1}v+C
\displaystyle \Rightarrow \log|vx|=\tan^{-1}v+C
\displaystyle \Rightarrow |vx|=Ce^{\tan^{-1}v}
\displaystyle \Rightarrow |y|=Ce^{\tan^{-1}\!\left(\frac{y}{x}\right)}

\displaystyle \textbf{Question 21: }~\left[x\sqrt{x^{2}+y^{2}}-y^{2}\right]dx+xy\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \left[x\sqrt{x^{2}+y^{2}}-y^{2}\right]dx+xy\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y^{2}-x\sqrt{x^{2}+y^{2}}}{xy}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-x^{2}\sqrt{1+v^{2}}}{vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v^{2}-\sqrt{1+v^{2}}}{v}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{\sqrt{1+v^{2}}}{v}
\displaystyle \Rightarrow \frac{v}{\sqrt{1+v^{2}}}\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Putting } 1+v^{2}=t
\displaystyle \Rightarrow v\,dv=\frac{1}{2}\,dt
\displaystyle \Rightarrow \frac{1}{2\sqrt{t}}\,dt=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{2\sqrt{t}}\,dt=-\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \sqrt{t}=-\log|x|+C
\displaystyle \Rightarrow \sqrt{1+v^{2}}=\log\!\left|\frac{C}{x}\right|
\displaystyle \Rightarrow \sqrt{1+\left(\frac{y}{x}\right)^{2}}=\log\!\left|\frac{C}{x}\right|
\displaystyle \Rightarrow \sqrt{x^{2}+y^{2}}=x\log\!\left|\frac{C}{x}\right|

\displaystyle \textbf{Question 22: }~x\frac{dy}{dx}=y-x\cos^{2}\left(\frac{y}{x}\right).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\frac{dy}{dx}=y-x\cos^{2}\!\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}-\cos^{2}\!\left(\frac{y}{x}\right)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=v-\cos^{2}v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\cos^{2}v
\displaystyle \Rightarrow \frac{1}{\cos^{2}v}\,dv=-\frac{1}{x}\,dx
\displaystyle \Rightarrow \sec^{2}v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \sec^{2}v\,dv=-\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \tan v=-\log|x|+C
\displaystyle \Rightarrow \tan v=\log\!\left|\frac{C}{x}\right|
\displaystyle \Rightarrow \tan\!\left(\frac{y}{x}\right)=\log\!\left|\frac{C}{x}\right|

\displaystyle \textbf{Question 23: }~\frac{y}{x}\cos\left(\frac{y}{x}\right)\,dx-\left[\frac{x}{y}\sin\left(\frac{y}{x}\right)+\cos\left(\frac{y}{x}\right)\right]dy=0.
\displaystyle \text{Answer:}
\displaystyle \frac{y}{x}\cos\!\left(\frac{y}{x}\right)\,dx-\left(\frac{x}{y}\sin\!\left(\frac{y}{x}\right)+\cos\!\left(\frac{y}{x}\right)\right)dy=0
\displaystyle \Rightarrow \left(\frac{x}{y}\sin\!\left(\frac{y}{x}\right)+\cos\!\left(\frac{y}{x}\right)\right)dy=\frac{y}{x}\cos\!\left(\frac{y}{x}\right)\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{\frac{y}{x}\cos\!\left(\frac{y}{x}\right)}{\frac{x}{y}\sin\!\left(\frac{y}{x}\right)+\cos\!\left(\frac{y}{x}\right)}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx}, \text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{v\cos v}{\frac{1}{v}\sin v+\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\cos v}{\frac{1}{v}\sin v+\cos v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{\sin v}{\frac{1}{v}\sin v+\cos v}
\displaystyle \Rightarrow \frac{\frac{1}{v}\sin v+\cos v}{\sin v}\,dv=-\frac{1}{x}\,dx
\displaystyle \Rightarrow \left(\frac{1}{v}+\cot v\right)dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \left(\frac{1}{v}+\cot v\right)dv=-\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \int \frac{1}{v}\,dv+\int \cot v\,dv=-\int \frac{1}{x}\,dx
\displaystyle \Rightarrow \log|v|+\log|\sin v|=-\log|x|+C
\displaystyle \Rightarrow \log|v\sin v|=-\log|x|+C
\displaystyle \Rightarrow \log|vx\sin v|=C
\displaystyle \Rightarrow |vx\sin v|=C
\displaystyle \Rightarrow |y\sin\!\left(\frac{y}{x}\right)|=C

\displaystyle \textbf{Question 24: }~xy\log\left(\frac{x}{y}\right)\,dx+\left[y^{2}-x^{2}\log\left(\frac{x}{y}\right)\right]dy=0. \hspace{2.0cm} \text{[CBSE2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Given:}
\displaystyle xy\log\!\left(\frac{x}{y}\right)\,dx+\left\{y^{2}-x^{2}\log\!\left(\frac{x}{y}\right)\right\}dy=0
\displaystyle \Rightarrow xy\log\!\left(\frac{x}{y}\right)\,dx=-\left\{y^{2}-x^{2}\log\!\left(\frac{x}{y}\right)\right\}dy
\displaystyle \Rightarrow \frac{dx}{dy}=\frac{x^{2}\log\!\left(\frac{x}{y}\right)-y^{2}}{xy\log\!\left(\frac{x}{y}\right)}
\displaystyle \text{It is a homogeneous differential equation.}
\displaystyle \text{Putting } x=vy \text{ and } \frac{dx}{dy}=v+y\frac{dv}{dy}, \text{ we get}
\displaystyle v+y\frac{dv}{dy}=\frac{v^{2}y^{2}\log v-y^{2}}{vy^{2}\log v}
\displaystyle \Rightarrow v+y\frac{dv}{dy}=\frac{v^{2}\log v-1}{v\log v}
\displaystyle \Rightarrow y\frac{dv}{dy}=\frac{v^{2}\log v-1}{v\log v}-v
\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{1}{v\log v}
\displaystyle \Rightarrow v\log v\,dv=-\frac{1}{y}\,dy
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int v\log v\,dv=-\int \frac{1}{y}\,dy
\displaystyle \Rightarrow \frac{v^{2}}{2}\log v-\frac{v^{2}}{4}=-\log|y|+C
\displaystyle \Rightarrow \frac{v^{2}}{2}\left(\log v-\frac{1}{2}\right)=-\log|y|+C
\displaystyle \Rightarrow v^{2}\left(\log v-\frac{1}{2}\right)=-2\log|y|+C
\displaystyle \Rightarrow \frac{x^{2}}{y^{2}}\left(\log\!\left(\frac{x}{y}\right)-\frac{1}{2}\right)+\log y^{2}=C

\displaystyle \textbf{Question 25: }~(1+e^{x/y})\,dx+e^{x/y}\left(1-\frac{x}{y}\right)dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \left(1+e^{x/y}\right)dx+e^{x/y}\left(1-\frac{x}{y}\right)dy=0
\displaystyle \Rightarrow \frac{dx}{dy}=-\frac{e^{x/y}\left(1-\frac{x}{y}\right)}{1+e^{x/y}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } x=vy \text{ and } \frac{dx}{dy}=v+y\frac{dv}{dy}, \text{ we get}
\displaystyle v+y\frac{dv}{dy}=-\frac{e^{v}(1-v)}{1+e^{v}}
\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{e^{v}(1-v)}{1+e^{v}}-v
\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{e^{v}-ve^{v}}{1+e^{v}}-\frac{v(1+e^{v})}{1+e^{v}}
\displaystyle \Rightarrow y\frac{dv}{dy}=-\frac{e^{v}+v}{1+e^{v}}
\displaystyle \Rightarrow \frac{1+e^{v}}{v+e^{v}}\,dv=-\frac{1}{y}\,dy
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1+e^{v}}{v+e^{v}}\,dv=-\int \frac{1}{y}\,dy
\displaystyle \Rightarrow \log|v+e^{v}|=-\log|y|+C
\displaystyle \Rightarrow \log|v+e^{v}|+\log|y|=C
\displaystyle \Rightarrow \log|y(v+e^{v})|=C
\displaystyle \Rightarrow y(v+e^{v})=C
\displaystyle \Rightarrow y\left(\frac{x}{y}+e^{x/y}\right)=C
\displaystyle \Rightarrow x+ye^{x/y}=C

\displaystyle \textbf{Question 26: }~(x^{2}+y^{2})\frac{dy}{dx}=8x^{2}-3xy+2y^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{2}+y^{2})\frac{dy}{dx}=8x^{2}-3xy+2y^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{8x^{2}-3xy+2y^{2}}{x^{2}+y^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{8x^{2}-3vx^{2}+2v^{2}x^{2}}{x^{2}+v^{2}x^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{8-3v+2v^{2}}{1+v^{2}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{8-4v+2v^{2}-v^{3}}{1+v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{4(2-v)+v^{2}(2-v)}{1+v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{(4+v^{2})(2-v)}{1+v^{2}}
\displaystyle \Rightarrow \frac{1+v^{2}}{(4+v^{2})(2-v)}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{1+v^{2}}{(4+v^{2})(2-v)}\,dv=\int\frac{1}{x}\,dx\ldots\ldots(1)
\displaystyle \text{Let us consider the left hand side of (1).}
\displaystyle \text{Using partial fraction,}
\displaystyle \text{Let } \frac{1+v^{2}}{(4+v^{2})(2-v)}=\frac{Av+B}{4+v^{2}}+\frac{C}{2-v}
\displaystyle \Rightarrow 1+v^{2}=Av(2-v)+B(2-v)+C(4+v^{2})
\displaystyle \Rightarrow 1+v^{2}=2Av-Av^{2}+2B-Bv+4C+Cv^{2}
\displaystyle \text{Comparing the coefficients of both sides, we get}
\displaystyle 2A-B=0
\displaystyle -A+C=1
\displaystyle 2B+4C=1
\displaystyle \text{Solving these three equations, we get}
\displaystyle A=-\frac{3}{8},\;B=-\frac{3}{4}\text{ and }C=\frac{5}{8}
\displaystyle \therefore \frac{1+v^{2}}{(4+v^{2})(2-v)}=\frac{-\frac{3}{8}v-\frac{3}{4}}{4+v^{2}}+\frac{\frac{5}{8}}{2-v}\ldots\ldots(2)
\displaystyle \text{From (1) and (2), we get}
\displaystyle \int\left(\frac{-\frac{3}{8}v-\frac{3}{4}}{4+v^{2}}+\frac{\frac{5}{8}}{2-v}\right)\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{3}{8}\int\frac{v}{v^{2}+4}\,dv-\frac{3}{4}\int\frac{1}{v^{2}+4}\,dv+\frac{5}{8}\int\frac{1}{2-v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{3}{16}\log\left|v^{2}+4\right|-\frac{3}{8}\tan^{-1}\!\left(\frac{v}{2}\right)-\frac{5}{8}\log\left|2-v\right|=\log|x|+\log C
\displaystyle \Rightarrow -\frac{3}{8}\tan^{-1}\!\left(\frac{v}{2}\right)=\log\!\left(Cx(2-v)^{\frac{5}{8}}(v^{2}+4)^{\frac{3}{16}}\right)
\displaystyle \Rightarrow e^{-\frac{3}{8}\tan^{-1}\left(\frac{v}{2}\right)}=C\left|x(2-v)^{\frac{5}{8}}(v^{2}+4)^{\frac{3}{16}}\right|
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow e^{-\frac{3}{8}\tan^{-1}\left(\frac{y}{2x}\right)}=C\left|x\left(2-\frac{y}{x}\right)^{\frac{5}{8}}\left(\frac{y^{2}}{x^{2}}+4\right)^{\frac{3}{16}}\right|
\displaystyle \Rightarrow e^{-\frac{3}{8}\tan^{-1}\left(\frac{y}{2x}\right)}=C\left|x\cdot\frac{1}{x}(2x-y)^{\frac{5}{8}}(y^{2}+4x^{2})^{\frac{3}{16}}\right|
\displaystyle \Rightarrow e^{-\frac{3}{8}\tan^{-1}\left(\frac{y}{2x}\right)}=C\left|2x-y\right|^{\frac{5}{8}}(y^{2}+4x^{2})^{\frac{3}{16}}
\displaystyle \text{Hence, } e^{-\frac{3}{8}\tan^{-1}\left(\frac{y}{2x}\right)}=C\left|2x-y\right|^{\frac{5}{8}}(y^{2}+4x^{2})^{\frac{3}{16}}\text{ is the required solution.}

\displaystyle \textbf{Question 27: }~(x^{2}-2xy)\,dy+(x^{2}-3xy+2y^{2})\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{2}-2xy)\,dy+(x^{2}-3xy+2y^{2})\,dx=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}-3xy+2y^{2}}{2xy-x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}-3vx^{2}+2v^{2}x^{2}}{2vx^{2}-x^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1-3v+2v^{2}}{2v-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-3v+2v^{2}}{2v-1}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1-2v}{2v-1}
\displaystyle \Rightarrow x\frac{dv}{dx}=-1
\displaystyle \Rightarrow dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow v=-\log|x|+C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \frac{y}{x}=-\log|x|+C
\displaystyle \Rightarrow \frac{y}{x}+\log|x|=C
\displaystyle \text{Hence, } \frac{y}{x}+\log|x|=C \text{ is the required solution.}

\displaystyle \textbf{Question 28: }~x\frac{dy}{dx}=y-x\cos^{2}\left(\frac{y}{x}\right).
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}=y-x\cos^{2}\!\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y-x\cos^{2}\!\left(\frac{y}{x}\right)}{x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx-x\cos^{2}v}{x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=v-\cos^{2}v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\cos^{2}v
\displaystyle \Rightarrow \sec^{2}v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\sec^{2}v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \tan v=-\log|x|+\log C
\displaystyle \Rightarrow \tan v=\log\left|\frac{C}{x}\right|
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \tan\!\left(\frac{y}{x}\right)=\log\left|\frac{C}{x}\right|
\displaystyle \text{Hence, } \tan\!\left(\frac{y}{x}\right)=\log\left|\frac{C}{x}\right|\text{ is the required solution.}

\displaystyle \textbf{Question 29: }~x\frac{dy}{dx}-y=2\sqrt{y^{2}-x^{2}}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\frac{dy}{dx}-y=2\sqrt{y^{2}-x^{2}}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\sqrt{y^{2}-x^{2}}+y}{x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{2\sqrt{v^{2}x^{2}-x^{2}}+vx}{x}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=2\sqrt{v^{2}-1}+v
\displaystyle \Rightarrow x\frac{dv}{dx}=2\sqrt{v^{2}-1}
\displaystyle \Rightarrow \frac{1}{2\sqrt{v^{2}-1}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{1}{2\sqrt{v^{2}-1}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{1}{\sqrt{v^{2}-1}}\,dv=2\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log\left|v+\sqrt{v^{2}-1}\right|=2\log|x|+\log C
\displaystyle \Rightarrow v+\sqrt{v^{2}-1}=Cx^{2}
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}-1}=Cx^{2}
\displaystyle \Rightarrow y+\sqrt{y^{2}-x^{2}}=Cx^{3}
\displaystyle \text{Hence, } y+\sqrt{y^{2}-x^{2}}=Cx^{3}\text{ is the required solution.}

\displaystyle \textbf{Question 30: }~x\cos\left(\frac{y}{x}\right)(y\,dx+x\,dy)=y\sin\left(\frac{y}{x}\right)(x\,dy-y\,dx). \hspace{2.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\cos\!\left(\frac{y}{x}\right)(y\,dx+x\,dy)=y\sin\!\left(\frac{y}{x}\right)(x\,dy-y\,dx)
\displaystyle \Rightarrow xy\cos\!\left(\frac{y}{x}\right)\,dx+x^{2}\cos\!\left(\frac{y}{x}\right)\,dy=xy\sin\!\left(\frac{y}{x}\right)\,dy-y^{2}\sin\!\left(\frac{y}{x}\right)\,dx
\displaystyle \Rightarrow \left[xy\cos\!\left(\frac{y}{x}\right)+y^{2}\sin\!\left(\frac{y}{x}\right)\right]dx=\left[xy\sin\!\left(\frac{y}{x}\right)-x^{2}\cos\!\left(\frac{y}{x}\right)\right]dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{xy\cos\!\left(\frac{y}{x}\right)+y^{2}\sin\!\left(\frac{y}{x}\right)}{xy\sin\!\left(\frac{y}{x}\right)-x^{2}\cos\!\left(\frac{y}{x}\right)}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx^{2}\cos v+v^{2}x^{2}\sin v}{vx^{2}\sin v-x^{2}\cos v}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+v^{2}\sin v}{v\sin v-\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\cos v+v^{2}\sin v}{v\sin v-\cos v}-v
\displaystyle \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}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}
\displaystyle \Rightarrow \frac{v\sin v-\cos v}{2v\cos v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{v\sin v-\cos v}{2v\cos v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{v\sin v-\cos v}{v\cos v}\,dv=2\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\tan v\,dv-\int\frac{1}{v}\,dv=2\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log|\sec v|-\log|v|=2\log|x|+\log C
\displaystyle \Rightarrow \log\left|\frac{\sec v}{v}\right|=\log\left|Cx^{2}\right|
\displaystyle \Rightarrow \frac{\sec v}{v}=Cx^{2}
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \sec\!\left(\frac{y}{x}\right)=Cx y
\displaystyle \text{Hence, } \sec\!\left(\frac{y}{x}\right)=Cxy\text{ is the required solution.}

\displaystyle \textbf{Question 31: }~(x^{2}+3xy+y^{2})\,dx-x^{2}\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{2}+3xy+y^{2})\,dx-x^{2}\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}+3xy+y^{2}}{x^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}+3vx^{2}+v^{2}x^{2}}{x^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=1+3v+v^{2}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=1+2v+v^{2}
\displaystyle \Rightarrow \frac{1}{1+2v+v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{1}{1+2v+v^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{1}{(1+v)^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{1}{1+v}=\log|x|+\log C
\displaystyle \Rightarrow \log|x|+\frac{1}{1+v}=-\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \log|x|+\frac{x}{x+y}=C_{1}
\displaystyle \text{where } C_{1}=-\log C
\displaystyle \text{Hence, } \log|x|+\frac{x}{x+y}=C_{1}\text{ is the required solution.}

\displaystyle \textbf{Question 32: }~(x-y)\frac{dy}{dx}=x+2y. \hspace{2.0cm} \text{[CBSE 2010, 2015]}
\displaystyle \text{Answer:}
\displaystyle (x-y)\frac{dy}{dx}=x+2y
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+2vx}{x-vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+2v}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{1-v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^{2}}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v+v^{2}}{1-v}
\displaystyle \Rightarrow \frac{1-v}{1+v+v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{1-v}{1+v+v^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{-(v-1)}{v^{2}+v+1}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{1}{2}\cdot\frac{2v-2}{v^{2}+v+1}\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{2v+1-3}{v^{2}+v+1}\,dv=-\int\frac{2}{x}\,dx
\displaystyle \Rightarrow \int\frac{2v+1}{v^{2}+v+1}\,dv-\int\frac{3}{v^{2}+v+1}\,dv=-\int\frac{2}{x}\,dx
\displaystyle \text{Let } I_{1}=\int\frac{2v+1}{v^{2}+v+1}\,dv
\displaystyle \text{and } I_{2}=\int\frac{3}{v^{2}+v+1}\,dv
\displaystyle I=I_{1}+I_{2}
\displaystyle I_{1}=\int\frac{2v+1}{v^{2}+v+1}\,dv
\displaystyle \text{Let } v^{2}+v+1=t \Rightarrow (2v+1)\,dv=dt
\displaystyle \Rightarrow I_{1}=\int\frac{dt}{t}=\log|t|=\log|v^{2}+v+1|
\displaystyle \Rightarrow I_{1}=\log|v^{2}+v+1|
\displaystyle I_{2}=\int\frac{3}{v^{2}+v+1}\,dv
\displaystyle \Rightarrow I_{2}=\int\frac{3}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dv
\displaystyle \Rightarrow I_{2}=3\cdot\frac{2}{\sqrt{3}}\tan^{-1}\!\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=2\sqrt{3}\tan^{-1}\!\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)
\displaystyle \Rightarrow I=\log|v^{2}+v+1|+2\sqrt{3}\tan^{-1}\!\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)
\displaystyle \Rightarrow \log|v^{2}+v+1|+2\sqrt{3}\tan^{-1}\!\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=-2\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \log\left|x^{2}+xy+y^{2}\right|=2\sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{x\sqrt{3}}\right)+\log C

\displaystyle \textbf{Solve the following differential equations:}

\displaystyle \textbf{Question 33: }~(2x^{2}y+y^{3})\,dx+(xy^{2}-3x^{3})\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (2x^{2}y+y^{3})\,dx+(xy^{2}-3x^{3})\,dy=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2x^{2}y+y^{3}}{3x^{3}-xy^{2}}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{2vx^{3}+v^{3}x^{3}}{3x^{3}-v^{2}x^{3}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{2v+v^{3}}{3-v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2v+v^{3}}{3-v^{2}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2v+v^{3}-3v+v^{3}}{3-v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{2v^{3}-v}{3-v^{2}}
\displaystyle \Rightarrow \frac{3-v^{2}}{2v^{3}-v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{3-v^{2}}{2v^{3}-v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow 3\int\frac{1}{2v^{3}-v}\,dv-\int\frac{v^{2}}{2v^{3}-v}\,dv=\int\frac{1}{x}\,dx\ldots\ldots(1)
\displaystyle \text{Considering } \frac{1}{2v^{3}-v}=\frac{1}{v(2v^{2}-1)},
\displaystyle \text{Let } \frac{1}{v(2v^{2}-1)}=\frac{A}{v}+\frac{Bv+C}{2v^{2}-1}\ldots\ldots(2)
\displaystyle 1=A(2v^{2}-1)+(Bv+C)v
\displaystyle \Rightarrow 1=2Av^{2}-A+Bv^{2}+Cv
\displaystyle \text{Comparing the coefficients of both sides, we get}
\displaystyle 2A+B=0
\displaystyle C=0
\displaystyle A=-1
\displaystyle \Rightarrow -2+B=0
\displaystyle \Rightarrow B=2
\displaystyle \text{Substituting } A=-1,\;B=2,\;C=0 \text{ in (2), we get}
\displaystyle \frac{1}{v(2v^{2}-1)}=-\frac{1}{v}+\frac{2v}{2v^{2}-1}\ldots\ldots(3)
\displaystyle \text{From (1) and (3), we get}
\displaystyle 3\int\left(-\frac{1}{v}+\frac{2v}{2v^{2}-1}\right)dv-\int\frac{v^{2}}{2v^{3}-v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -3\int\frac{1}{v}\,dv+5\int\frac{v}{2v^{2}-1}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -3\log|v|+\frac{5}{4}\log|2v^{2}-1|=\log|x|+\log C
\displaystyle \Rightarrow 12\log\left|\frac{1}{v}\right|+5\log|2v^{2}-1|=4\log|Cx|
\displaystyle \Rightarrow \log\left|\frac{1}{v^{12}}(2v^{2}-1)^{5}\right|=\log\left|C^{4}x^{4}\right|
\displaystyle \Rightarrow \left|\frac{1}{v^{12}}(2v^{2}-1)^{5}\right|=\left|C^{4}x^{4}\right|
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \left|\frac{x^{12}}{y^{12}}\left(\frac{2y^{2}}{x^{2}}-1\right)^{5}\right|=\left|C^{4}x^{4}\right|
\displaystyle \Rightarrow \left|\frac{(2y^{2}-x^{2})^{5}}{x^{10}}\right|=\left|C^{4}x^{4}\right|
\displaystyle \Rightarrow C^{4}x^{2}y^{12}=\left|2y^{2}-x^{2}\right|^{5}
\displaystyle \text{Hence, } C^{4}x^{2}y^{12}=\left|2y^{2}-x^{2}\right|^{5}\text{ is the required solution.}

\displaystyle \textbf{Question 34: }~x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\frac{dy}{dx}-y+x\sin\!\left(\frac{y}{x}\right)=0
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y-x\sin\!\left(\frac{y}{x}\right)}{x}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx-x\sin v}{x}
\displaystyle \Rightarrow x\frac{dv}{dx}=v-\sin v-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\sin v
\displaystyle \Rightarrow \mathrm{cosec}\,v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \mathrm{cosec}\,v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\int \mathrm{cosec}\,v\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\log\left|\mathrm{cosec}\,v-\cot v\right|=\log|x|+\log C
\displaystyle \Rightarrow \log\left|\frac{1}{\mathrm{cosec}\,v-\cot v}\right|=\log|Cx|
\displaystyle \Rightarrow \log\left|\mathrm{cosec}\,v+\cot v\right|=\log|Cx|
\displaystyle \Rightarrow \log\left|\frac{1+\cos v}{\sin v}\right|=\log|Cx|
\displaystyle \Rightarrow \frac{1+\cos v}{\sin v}=Cx
\displaystyle \Rightarrow x\sin v=\frac{1}{C}(1+\cos v)
\displaystyle \Rightarrow x\sin v=K(1+\cos v)\text{ where }K=\frac{1}{C}
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow x\sin\!\left(\frac{y}{x}\right)=K\left[1+\cos\!\left(\frac{y}{x}\right)\right]
\displaystyle \text{Hence, } x\sin\!\left(\frac{y}{x}\right)=K\left[1+\cos\!\left(\frac{y}{x}\right)\right]\text{ is the required solution.}

\displaystyle \textbf{Question 35: }~y\,dx+\left[x\log\left(\frac{y}{x}\right)\right]dy-2x\,dy=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle y\,dx+\left\{x\log\!\left(\frac{y}{x}\right)\right\}dy-2x\,dy=0
\displaystyle \Rightarrow \left\{2x-x\log\!\left(\frac{y}{x}\right)\right\}dy=y\,dx
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{2x-x\log\!\left(\frac{y}{x}\right)}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx}{2x-x\log v}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v}{2-\log v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v}{2-\log v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v-2v+v\log v}{2-\log v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\log v-v}{2-\log v}
\displaystyle \Rightarrow \frac{2-\log v}{v\log v-v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{2-\log v}{v\log v-v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{1-(\log v-1)}{v(\log v-1)}\,dv=\int\frac{1}{x}\,dx
\displaystyle \text{Putting } \log v-1=t
\displaystyle \Rightarrow \frac{1}{v}\,dv=dt
\displaystyle \Rightarrow \int\frac{1-t}{t}\,dt=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\left(\frac{1}{t}-1\right)dt=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log|t|-t=\log|x|+\log C
\displaystyle \Rightarrow \log|\log v-1|-(\log v-1)=\log|x|+\log C
\displaystyle \Rightarrow \log|\log v-1|-\log v=\log|x|+\log C_{1}
\displaystyle \Rightarrow \log\left|\frac{\log v-1}{v}\right|=\log|C_{1}x|
\displaystyle \Rightarrow \frac{\log v-1}{v}=C_{1}x
\displaystyle \Rightarrow \log v-1=C_{1}xv
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \log\! |\frac{y}{x}|-1=C_{1}y
\displaystyle \text{Hence, } \log\! |\frac{y}{x}|-1=C_{1}y\text{ is the required solution.}

\displaystyle \textbf{Question 36: }~\text{Solve each of the following initial value problems:}

\displaystyle \text{(i)}~(x^{2}+y^{2})\,dx=2xy\,dy,\ y(1)=0.
\displaystyle \text{Answer:}
\displaystyle (x^{2}+y^{2})\,dx=2xy\,dy,\;y(1)=0
\displaystyle \text{We have,}
\displaystyle (x^{2}+y^{2})\,dx=2xy\,dy\ldots\ldots(1)
\displaystyle \text{This is a homogeneous equation, so let us take } y=vx
\displaystyle \text{Then, } \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Putting } y=vx \text{ in equation (1)}
\displaystyle (x^{2}+v^{2}x^{2})=2vx^{2}\left(v+x\frac{dv}{dx}\right)
\displaystyle x^{2}(1+v^{2})=2vx^{2}\left(v+x\frac{dv}{dx}\right)
\displaystyle \Rightarrow 1+v^{2}=2v^{2}+2vx\frac{dv}{dx}
\displaystyle \Rightarrow 1-v^{2}=2vx\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dx}{x}=\frac{2v}{1-v^{2}}\,dv
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{1}{x}\,dx=\int\frac{2v}{1-v^{2}}\,dv
\displaystyle \text{Let } (1-v^{2})=t
\displaystyle \Rightarrow -2v\,dv=dt
\displaystyle \Rightarrow \log x=-\int\frac{dt}{t}
\displaystyle \Rightarrow \log x=-\log t+\log C
\displaystyle \Rightarrow \log x=-\log(1-v^{2})+\log C
\displaystyle \Rightarrow \log\left(\frac{x}{1-v^{2}}\right)=\log C
\displaystyle \Rightarrow \log\left(\frac{x^{2}}{x^{2}-y^{2}}\right)=\log C
\displaystyle \text{As } y(1)=0
\displaystyle \Rightarrow \log\left(\frac{1}{1}\right)=\log C
\displaystyle \Rightarrow C=1
\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{x}=1
\displaystyle \Rightarrow x^{2}-y^{2}=x

\displaystyle \text{(ii)}~xe^{y/x}-y+x\frac{dy}{dx}=0,\ y(e)=0.
\displaystyle \text{Answer:}
\displaystyle x e^{\frac{y}{x}}-y+x\frac{dy}{dx}=0,\;y(e)=0
\displaystyle \text{This is also a homogeneous equation.}
\displaystyle \text{Put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle x e^{v}-vx+x\left(v+x\frac{dv}{dx}\right)=0
\displaystyle \Rightarrow x e^{v}-vx+xv+x^{2}\frac{dv}{dx}=0
\displaystyle \Rightarrow x e^{v}+x^{2}\frac{dv}{dx}=0
\displaystyle \Rightarrow e^{v}+x\frac{dv}{dx}=0
\displaystyle \Rightarrow e^{v}=-x\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dx}{x}=-\frac{1}{e^{v}}\,dv
\displaystyle \text{On integration both sides we get}
\displaystyle \int\frac{1}{x}\,dx=-\int e^{-v}\,dv
\displaystyle \Rightarrow \log x=e^{-v}+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \log x=e^{-\frac{y}{x}}+\log C
\displaystyle \text{As given } y(e)=0
\displaystyle \Rightarrow \log e=e^{0}+\log C
\displaystyle \Rightarrow 1=1+\log C
\displaystyle \Rightarrow \log C=0
\displaystyle \Rightarrow \log x=e^{-\frac{y}{x}}

\displaystyle \text{(iii)}~\frac{dy}{dx}-\frac{y}{x}+\mathrm{cosec}\left(\frac{y}{x}\right)=0,\ y(1)=0. \hspace{2.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}-\frac{y}{x}+\mathrm{cosec}\!\left(\frac{y}{x}\right)=0,\;y(1)=0
\displaystyle \text{This is a homogeneous equation, put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}-v+\mathrm{cosec}\,v=0
\displaystyle \Rightarrow x\frac{dv}{dx}=-\mathrm{cosec}\,v
\displaystyle \Rightarrow \frac{dv}{\mathrm{cosec}\,v}=-\frac{1}{x}\,dx
\displaystyle \Rightarrow \sin v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\sin v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\cos v=-\log x+\log C
\displaystyle \Rightarrow -\cos v+\log x=\log C
\displaystyle \Rightarrow \cos v-\log x=-\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \cos\!\left(\frac{y}{x}\right)-\log x=-\log C
\displaystyle \text{As } y(1)=0
\displaystyle \Rightarrow \cos 0-\log 1=-\log C
\displaystyle \Rightarrow 1=-\log C
\displaystyle \Rightarrow \log C=-1
\displaystyle \Rightarrow \cos\!\left(\frac{y}{x}\right)+\log x=1

\displaystyle \text{(iv)}~(xy-y^{2})\,dx-x^{2}\,dy=0,\ y(1)=1.
\displaystyle \text{Answer:}
\displaystyle (xy-y^{2})\,dx-x^{2}\,dy=0,\;y(1)=1
\displaystyle \text{This is a homogeneous equation, put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle (xy-y^{2})=x^{2}\frac{dy}{dx}
\displaystyle (vx^{2}-v^{2}x^{2})=x^{2}\left(v+x\frac{dv}{dx}\right)
\displaystyle vx^{2}(1-v)=x^{2}\left(v+x\frac{dv}{dx}\right)
\displaystyle v(1-v)=v+x\frac{dv}{dx}
\displaystyle v-v^{2}=v+x\frac{dv}{dx}
\displaystyle -v^{2}=x\frac{dv}{dx}
\displaystyle -\frac{1}{x}\,dx=\frac{1}{v^{2}}\,dv
\displaystyle \text{On integrating both sides we get}
\displaystyle -\int\frac{1}{x}\,dx=\int\frac{1}{v^{2}}\,dv
\displaystyle \Rightarrow -\log x=\frac{v^{-1}}{-1}+\log C
\displaystyle \Rightarrow -\log x=-\frac{1}{v}+\log C
\displaystyle \Rightarrow \frac{1}{v}-\log x=\log C
\displaystyle \Rightarrow \frac{x}{y}-\log x=\log C
\displaystyle \text{As } y(1)=1
\displaystyle \Rightarrow 1-\log 1=\log C
\displaystyle \Rightarrow \log C=1
\displaystyle \Rightarrow \frac{x}{y}-\log x=1

\displaystyle \text{(v)}~\frac{dy}{dx}=\frac{y(x+2y)}{x(2x+y)},\ y(1)=2.
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\frac{y(x+2y)}{x(2x+y)},\;y(1)=1
\displaystyle \text{This is a homogeneous equation.}
\displaystyle \text{Put } y=vx
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v(x+2vx)}{2x+vx}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v(1+2v)}{2+v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v(1+2v)}{2+v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v+2v^{2}-2v-v^{2}}{2+v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v^{2}-v}{2+v}
\displaystyle \Rightarrow \frac{2+v}{v^{2}-v}\,dv=\frac{1}{x}\,dx
\displaystyle \text{On integrating both sides we get}
\displaystyle \int\frac{2+v}{v^{2}-v}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\left(\frac{2}{v(v-1)}+\frac{v}{v(v-1)}\right)dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow 2\int\left(\frac{1}{v-1}-\frac{1}{v}\right)dv+\int\frac{1}{v-1}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow 2\left(\log|v-1|-\log|v|\right)+\log|v-1|=\log|x|+\log C
\displaystyle \Rightarrow 2\log\left|\frac{v-1}{v}\right|+\log|v-1|=\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow 2\log\left|\frac{y-x}{y}\right|+\log\left|\frac{y-x}{x}\right|=\log|x|+\log C
\displaystyle \text{As } y(1)=1
\displaystyle \Rightarrow 2\log\left|\frac{1-1}{1}\right|+\log\left|\frac{1-1}{1}\right|=\log 1+\log C
\displaystyle \Rightarrow 0=0+\log C
\displaystyle \Rightarrow \log C=0
\displaystyle \Rightarrow 2\log\left|\frac{y-x}{y}\right|+\log\left|\frac{y-x}{x}\right|=\log x

\displaystyle \text{(vi)}~(y^{4}-2x^{3}y)\,dx+(x^{4}-2xy^{3})\,dy=0,\ y(1)=1.
\displaystyle \text{Answer:}
\displaystyle (y^{4}-2x^{3}y)\,dx+(x^{4}-2xy^{3})\,dy=0,\;y(1)=1
\displaystyle \text{This is a homogeneous equation, put } y=vx
\displaystyle v^{4}x^{4}-2vx^{4}+(x^{4}-2v^{3}x^{4})\left(v+x\frac{dv}{dx}\right)=0
\displaystyle \Rightarrow (v^{4}x^{4}-2vx^{4})=(2v^{3}x^{4}-x^{4})\left(v+x\frac{dv}{dx}\right)
\displaystyle \Rightarrow vx^{4}(v^{3}-2)=x^{4}(2v^{3}-1)\left(v+x\frac{dv}{dx}\right)
\displaystyle \Rightarrow v(v^{3}-2)=(2v^{3}-1)v+x(2v^{3}-1)\frac{dv}{dx}
\displaystyle \Rightarrow v(v^{3}-2-2v^{3}+1)=x(2v^{3}-1)\frac{dv}{dx}
\displaystyle \Rightarrow v(-1-v^{3})=x(2v^{3}-1)\frac{dv}{dx}
\displaystyle \Rightarrow v(1+v^{3})=x(1-2v^{3})\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dx}{x}=\frac{1-2v^{3}}{v(1+v^{3})}\,dv
\displaystyle \text{On integrating both sides we get}
\displaystyle \int\frac{1}{x}\,dx=\int\frac{1-2v^{3}}{v(1+v^{3})}\,dv
\displaystyle \Rightarrow \log x=\int\frac{1+v^{3}-3v^{3}}{v(1+v^{3})}\,dv
\displaystyle \Rightarrow \log x=\int\frac{1}{v}\,dv-\int\frac{3v^{2}}{1+v^{3}}\,dv
\displaystyle \Rightarrow \log x=\log v-\int\frac{dt}{t}\;\text{where }(1+v^{3})=t,\;3v^{2}dv=dt
\displaystyle \Rightarrow \log x=\log v-\log(1+v^{3})+\log C
\displaystyle \Rightarrow \log x=\log\left(\frac{v}{1+v^{3}}\right)+\log C
\displaystyle \text{As } v=\frac{y}{x}
\displaystyle \Rightarrow \log x=\log\left(\frac{\frac{y}{x}}{1+\frac{y^{3}}{x^{3}}}\right)+\log C
\displaystyle \Rightarrow \log x=\log\left(\frac{yx^{2}}{x^{3}+y^{3}}\right)+\log C
\displaystyle \text{As } y(1)=1
\displaystyle \Rightarrow \log 1=\log\left(\frac{1}{2}\right)+\log C
\displaystyle \Rightarrow 0=\log\left(\frac{1}{2}\right)+\log C
\displaystyle \Rightarrow \log C=\log 2
\displaystyle \Rightarrow \log x=\log\left(\frac{yx^{2}}{x^{3}+y^{3}}\right)+\log 2

\displaystyle \text{(vii)}~x(x^{2}+3y^{2})\,dx+y(y^{2}+3x^{2})\,dy=0,\ y(1)=1.
\displaystyle \text{Answer:}
\displaystyle x(x^{2}+3y^{2})\,dx+y(y^{2}+3x^{2})\,dy=0,\;y(1)=1
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{x(x^{2}+3y^{2})}{y(y^{2}+3x^{2})}
\displaystyle \text{This is a homogeneous equation. Put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=-\frac{x(x^{2}+3v^{2}x^{2})}{vx(v^{2}x^{2}+3x^{2})}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+3v^{2}}{v(v^{2}+3)}-3
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{4}+6v^{2}+1}{v(v^{2}+3)}
\displaystyle \Rightarrow \frac{v(v^{2}+3)}{v^{4}+6v^{2}+1}\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{v(v^{2}+3)}{v^{4}+6v^{2}+1}\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{4v^{3}+12v}{v^{4}+6v^{2}+1}\,dv=-4\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log\left|v^{4}+6v^{2}+1\right|=\log\left|\frac{C}{x^{4}}\right|
\displaystyle \Rightarrow \left|v^{4}+6v^{2}+1\right|=\left|\frac{C}{x^{4}}\right|
\displaystyle \Rightarrow \left|v^{4}x^{4}+6v^{2}x^{4}+x^{4}\right|=|C|\ldots\ldots(1)
\displaystyle \text{Putting } v=\frac{y}{x},\;x=1,\;y=1
\displaystyle \Rightarrow |1+6+1|=|C|
\displaystyle \Rightarrow C=8
\displaystyle \text{Putting } C=8 \text{ in (1),}
\displaystyle \Rightarrow y^{4}+x^{4}+6x^{2}y^{2}=8

\displaystyle \text{(viii)}~\left[x\sin^{2}\left(\frac{y}{x}\right)-y\right]dx+x\,dy=0,\ y(1)=\frac{\pi}{4}. \hspace{2.0cm} \text{[CBSE 2011, 2013]}
\displaystyle \text{Answer:}
\displaystyle x\sin^{2}\!\left(\frac{y}{x}\right)-y\,dx+x\,dy=0,\;y(1)=\frac{\pi}{4}
\displaystyle \text{This is a homogeneous equation, so put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=-\sin^{2}\!\left(\frac{vx}{x}\right)+\frac{vx}{x}
\displaystyle \Rightarrow x\frac{dv}{dx}=-\sin^{2}v
\displaystyle \Rightarrow \frac{dv}{\sin^{2}v}=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\csc^{2}v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\cot v=-\log|x|+\log C
\displaystyle \Rightarrow \cot v=\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \cot\!\left(\frac{y}{x}\right)=\log|x|+\log C
\displaystyle \text{Using } y(1)=\frac{\pi}{4}
\displaystyle \Rightarrow \cot\!\left(\frac{\pi}{4}\right)=\log 1+\log C
\displaystyle \Rightarrow 1=\log C
\displaystyle \Rightarrow C=e
\displaystyle \text{Hence, } \cot\!\left(\frac{y}{x}\right)=\log(ex)

\displaystyle \text{(ix)}~x\frac{dy}{dx}-y+x\sin\left(\frac{y}{x}\right)=0,\ y(2)=x. \hspace{2.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}-y+x\sin\!\left(\frac{y}{x}\right)=0,\;y(2)=\pi
\displaystyle \text{It is a homogeneous equation, put } y=vx
\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle v+x\frac{dv}{dx}=\frac{vx}{x}-\sin\!\left(\frac{vx}{x}\right)
\displaystyle \Rightarrow x\frac{dv}{dx}=-\sin v
\displaystyle \Rightarrow \frac{dv}{\sin v}=-\frac{1}{x}\,dx
\displaystyle \Rightarrow \mathrm{cosec}\,v\,dv=-\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides we get}
\displaystyle \int \mathrm{cosec}\,v\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log\left|\mathrm{cosec}\,v-\cot v\right|=-\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \log\left|\mathrm{cosec}\!\left(\frac{y}{x}\right)-\cot\!\left(\frac{y}{x}\right)\right|=-\log x+\log C
\displaystyle \text{Putting } x=2,\;y=\pi
\displaystyle \Rightarrow \log\left|\mathrm{cosec}\!\left(\frac{\pi}{2}\right)-\cot\!\left(\frac{\pi}{2}\right)\right|=-\log 2+\log C
\displaystyle \Rightarrow 0=-\log 2+\log C
\displaystyle \Rightarrow \log C=\log 2
\displaystyle \Rightarrow \log\left|\mathrm{cosec}\!\left(\frac{y}{x}\right)-\cot\!\left(\frac{y}{x}\right)\right|=-\log x+\log 2

\displaystyle \textbf{Question 37: }~\text{Find the particular solution of the differential equation }  \\ x\cos\left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos\left(\frac{y}{x}\right)+x, \text{ given that when } x=1,\ y=\frac{\pi}{4}. \hspace{2.0cm} \text{[CBSE 2013, 2017]}
\displaystyle \text{Answer:}
\displaystyle x\cos\!\left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos\!\left(\frac{y}{x}\right)+x
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y\cos\!\left(\frac{y}{x}\right)+x}{x\cos\!\left(\frac{y}{x}\right)}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{vx\cos v+x}{x\cos v}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{v\cos v+1}{\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v\cos v+1-v\cos v}{\cos v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1}{\cos v}
\displaystyle \Rightarrow \cos v\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\cos v\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \sin v=\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \sin\!\left(\frac{y}{x}\right)=\log|x|+\log C\ldots\ldots(1)
\displaystyle \text{At } x=1,\;y=\frac{\pi}{4}\text{ (Given)}
\displaystyle \Rightarrow \sin\!\left(\frac{\pi}{4}\right)=\log 1+\log C
\displaystyle \Rightarrow \frac{1}{\sqrt{2}}=\log C
\displaystyle \Rightarrow C=e^{\frac{1}{\sqrt{2}}}
\displaystyle \text{Putting this value of } C \text{ in (1), we get}
\displaystyle \Rightarrow \sin\!\left(\frac{y}{x}\right)=\log |x|+\frac{1}{\sqrt{2}}
\displaystyle \text{Hence, } \sin\!\left(\frac{y}{x}\right)=\log |x|+\frac{1}{\sqrt{2}}\text{ is the required solution.}

\displaystyle \textbf{Question 38: }~\text{Find the particular solution of the differential equation } \\ (x-y)\frac{dy}{dx}=x+2y, \text{ given that when } x=1,\ y=0. \hspace{2.0cm} \text{[CBSE 2014, 2015]}
\displaystyle \text{Answer:}
\displaystyle (x-y)\frac{dy}{dx}=x+2y
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx},\text{ we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x+2vx}{x-vx}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v}{1-v}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+2v-v+v^{2}}{1-v}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{1+v+v^{2}}{1-v}
\displaystyle \Rightarrow \frac{1-v}{1+v+v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int\frac{1-v}{1+v+v^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\left(\frac{-\frac{1}{2}(2v+1)}{1+v+v^{2}}+\frac{3}{2}\cdot\frac{1}{1+v+v^{2}}\right)dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{1}{2}\int\frac{2v+1}{1+v+v^{2}}\,dv+\frac{3}{2}\int\frac{1}{\left(v+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow -\frac{1}{2}\log|1+v+v^{2}|+\sqrt{3}\tan^{-1}\!\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=\log|x|+\log C
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{2v+1}{\sqrt{3}}\right)-\frac{1}{2}\log|1+v+v^{2}|=\log|x|+\log C
\displaystyle \text{Putting } v=\frac{y}{x},\text{ we get}
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{1}{2}\log\left|1+\frac{y}{x}+\frac{y^{2}}{x^{2}}\right|=\log|x|+\log C
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{1}{2}\log\left|\frac{x^{2}+xy+y^{2}}{x^{2}}\right|=\log|x|+\log C
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{1}{2}\log|x^{2}+xy+y^{2}|+\log|x|=\log|x|+\log C
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{1}{2}\log|x^{2}+xy+y^{2}|=\log C\ldots\ldots(1)
\displaystyle \text{At } x=1,\;y=0\text{ (Given)}
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right)-\frac{1}{2}\log 1=\log C
\displaystyle \Rightarrow \log C=\sqrt{3}\cdot\frac{\pi}{6}=\frac{\pi}{2\sqrt{3}}
\displaystyle \text{Substituting the value of } \log C \text{ in (1), we get}
\displaystyle \Rightarrow \sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{1}{2}\log|x^{2}+xy+y^{2}|=\frac{\pi}{2\sqrt{3}}
\displaystyle \Rightarrow \log|x^{2}+xy+y^{2}|=2\sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{\pi}{\sqrt{3}}
\displaystyle \text{Hence, } \log|x^{2}+xy+y^{2}|=2\sqrt{3}\tan^{-1}\!\left(\frac{x+2y}{\sqrt{3}x}\right)-\frac{\pi}{\sqrt{3}}\text{ is the required solution.}

\displaystyle \textbf{Question 39: }~\text{Find the particular solution of the differential equation } \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}, \text{ given that } y=1 \text{ when } x=0. \hspace{2.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \frac{dy}{dx}=\frac{xy}{x^{2}+y^{2}}\ldots\ldots(1)
\displaystyle \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}
\displaystyle \text{Substituting the value of } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx} \text{ in (1), we get}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{x^{2}v}{x^{2}+x^{2}v^{2}}
\displaystyle \Rightarrow x\frac{dv}{dx}=\frac{v}{1+v^{2}}-v
\displaystyle \Rightarrow x\frac{dv}{dx}=-\frac{v^{3}}{1+v^{2}}
\displaystyle \Rightarrow -\frac{1+v^{2}}{v^{3}}\,dv=\frac{1}{x}\,dx
\displaystyle \Rightarrow \int-\frac{1+v^{2}}{v^{3}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\left(-v^{-3}-v^{-1}\right)dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \frac{1}{2v^{2}}-\log|v|=\log|x|+\log C
\displaystyle \Rightarrow \frac{1}{2v^{2}}-\log\!\left(\frac{y}{x}\right)=\log|x|+\log C
\displaystyle \Rightarrow \frac{x^{2}}{2y^{2}}-\log\!\left(\frac{y}{x}\right)=\log|x|+\log C\ldots\ldots(2)
\displaystyle \text{Using the given condition, we get } C=0
\displaystyle \text{Substituting the value of } C \text{ in (2), we get}
\displaystyle \Rightarrow \frac{x^{2}}{2y^{2}}-\log\!\left(\frac{y}{x}\right)=\log x
\displaystyle \Rightarrow \frac{x^{2}}{2y^{2}}=\log x+\log\!\left(\frac{y}{x}\right)
\displaystyle \Rightarrow \frac{x^{2}}{2y^{2}}=\log y

\displaystyle \textbf{Question 40: }~\text{Show that the family of curves for which } \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}, \text{ is given by }  \\ x^{2}-y^{2}=Cx. \hspace{2.0cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{The given differential equation is } \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy}\ldots\ldots(1)
\displaystyle \text{This is a homogeneous differential equation.}
\displaystyle \text{Putting } y=vx \text{ and } \frac{dy}{dx}=v+x\frac{dv}{dx} \text{ in (1), we get}
\displaystyle v+x\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{2vx^{2}}
\displaystyle \Rightarrow v+x\frac{dv}{dx}=\frac{1+v^{2}}{2v}
\displaystyle \Rightarrow \frac{1+v^{2}}{2v}-v=x\frac{dv}{dx}
\displaystyle \Rightarrow \frac{1-v^{2}}{2v}=x\frac{dv}{dx}
\displaystyle \Rightarrow \frac{2v}{1-v^{2}}\,dv=\frac{1}{x}\,dx
\displaystyle \text{Integrating on both sides, we get}
\displaystyle \int\frac{2v}{1-v^{2}}\,dv=\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \int\frac{-2v}{v^{2}-1}\,dv=-\int\frac{1}{x}\,dx
\displaystyle \Rightarrow \log(1-v^{2})=-\log x+\log C
\displaystyle \Rightarrow \log(1-v^{2})+\log x=\log C
\displaystyle \Rightarrow \log\left[(1-v^{2})x\right]=\log C
\displaystyle \Rightarrow (1-v^{2})x=C
\displaystyle \Rightarrow \left(1-\frac{y^{2}}{x^{2}}\right)x=C
\displaystyle \Rightarrow x^{2}-y^{2}=Cx
\displaystyle \text{Thus, the family of curves for which } \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2xy} \text{ is given by } x^{2}-y^{2}=Cx.

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