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

$x=1\&y=0$

Now,

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

$\because$ The above equation is Homogenous Differential Equation,

$\therefore$ Let $y=vx$ ___(1) $\Rightarrow \frac{dy}{dx}=v+\frac{xdv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}\Rightarrow V+\frac{xdv}{dx}=\frac{x+2vx}{x-vx}$

$\Rightarrow \frac{xdv}{dx}+V=\frac{1+2V}{1-V}\Rightarrow \frac{xdv}{dx}=\frac{1+2v}{1-v}-v$

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

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

Integrating both sides, we have

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

$\Rightarrow \frac{1}{2}\int \frac{2v+1-3}{v^2+v+1}dv=-\log x+C$

$\Rightarrow \frac{1}{2}\frac{2v+1}{v^2+v+1}dv-\frac{3}{2}\int \frac{1}{v^2+v+1}dv=-\log x+C$

Let $v^2+v+1=t\Rightarrow (2v=1)dv=dt$

$\Rightarrow \frac{1}{2}\int \frac{dt}{t}-\frac{3}{2}\int \frac{1}{{(v+\frac{1}{2})}^2+\frac{3}{4}}dv=-\log x+C$

$\Rightarrow \frac{1}{2}\log |t|-\frac{3}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{v+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=-\log x+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\frac{x}{a}]$

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

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

For $x=1\&y=0$

$\frac{1}{2}\log |0+0+1|-\sqrt{3}{\tan }^{-1}\left(\frac{0+1}{\sqrt{3}}\right)=-\log 1+C\Rightarrow 0-\frac{\sqrt{3\pi }}{5}=C$

$\Rightarrow C=\frac{-\sqrt{3\pi }}{6}=\frac{-\pi }{2\sqrt{3}}$

$\Rightarrow \frac{1}{2}\log |\frac{y^2}{x^2}+\frac{y}{x}+1|-\sqrt{3}{\tan }^{-1}\left[\frac{\frac{2y}{x}+1}{\sqrt{3}}\right]=-\log x-\frac{\pi }{2\sqrt{3}}$