Question

Find the particular solution of the differential equation $(x-y)\frac{dy}{dx}=(x+2y)$, given that $y=0$ when $x=1$.

Answer 1

Given, $(x-y)\frac{dy}{dx}=x+2y$

By simplifying the above equation, we have

$\frac{dy}{dx}=\frac{x+2y}{x-y}$ ...$\left(i\right)$

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

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

= $\frac{\lambda (x+2y)}{\lambda (x-y)}$

= ${\lambda }^{0}F(x,y)$

$F(x,y)$ is homogeneous function and hence given differential equation is homogeneous.

Now, let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

Substituting these values in equation $\left(i\right)$, we have

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

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

By integrating both sides, we obtain

$\int \frac{1-v}{v^2+v+1}dv=\int \frac{dx}{x}...\left(ii\right)$

L.H.S. = $\int \frac{1-v}{v^2+v+1}dv$

$\text{Let}1-v=A(2v+1)+B=2Av+(A+B)$
Comparing coefficients of both sides, we obtain

$2A=-1,A+B=1$

$orA=-\frac{1}{2},B=\frac{3}{2}$

$\therefore \int \frac{1-v}{v^2+v+1}dv=\int \frac{-\frac{1}{2}(2v+1)+\frac{3}{2}}{v^2+v+1}dv$

$=-\frac{1}{2}\int \frac{2v+1}{v^2+v+1}dv+\frac{3}{2}\int \frac{dv}{v^2+v+1}$

$=-\frac{1}{2}\int \frac{2v+1}{v^2+v+1}dv+\frac{3}{2}\int \frac{dv}{{(v+\frac{1}{2})}^2+\frac{3}{4}}$

$=-\frac{1}{2}\log |v^2+v+1|+\frac{3}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)$

$\text{Now, susbstituting it in equation}\left(ii\right)\text{, we obtain}$

$-\frac{1}{2}\log |v^2+v+1|+\sqrt{3}{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)=\log x+C$

$\Rightarrow -\frac{1}{2}\log |x^2+xy+y^2|+\frac{1}{2}\log x^2+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\log x+C$

$\Rightarrow -\frac{1}{2}\log |x^2+xy+y^2|+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=C$

$\text{Now,}y=0\text{, when}x=1$

$\Rightarrow -\frac{1}{2}\log \left(1\right)+\sqrt{3}{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=C$

$\Rightarrow 0+\sqrt{3}{\tan }^{-1}\left(\tan \frac{\pi }{6}\right)=C$

$\Rightarrow C=\frac{\sqrt{3}}{6}\pi$

Thus, required particular solution is:

$-\frac{1}{2}\log (x^2+xy+y^2)+\sqrt{3}{\tan }^{-1}\left(\frac{2y+x}{\sqrt{3}x}\right)=\frac{\sqrt{3}}{6}\pi$