Question

The solution of $y-x{y}_1=2(x+y{y}_1),y\left(1\right)=1$ is

• A. ${\tan }^{-1}y/x+\log (x^2+y^2)=\log 2$
• B. ${\tan }^{-1}y/x+\log \frac{1}{2}(x^2+y^2)=\pi /4$
• C. ${\tan }^{-1}y/x+\log \frac{1}{3}(x^2+y^2)=0$
• D. none of these

Answer 1

Answer: (B)

${\tan }^{-1}y/x+\log \frac{1}{2}(x^2+y^2)=\pi /4$

$y-x{y}^{\prime }=2(x+y{y}^{\prime })$
Let $y=vx\Rightarrow \frac{dy}{dx}=vx\frac{dv}{dx}$
$\therefore -x(x\frac{dv}{dx}+v)+xv=2(x+x(x\frac{dv}{dx}+v)v)$
$\Rightarrow -x^2\frac{dv}{dx}=2(x+x(x\frac{dv}{dx}+v)v)$
$\Rightarrow \frac{dv}{dx}=-\frac{2(v^2+1)}{x(2v+1)}\Rightarrow \frac{\frac{dv}{dx}(2v+1)}{v^2+1}=-\frac{2}{x}$
Integrating both sides
$\int \frac{\frac{dv}{dx}(2v+1)}{v^2+1}dx=-\int \frac{2}{x}dx\Rightarrow {\tan }^{-1}v+\log (v^2+1)=-2\log x+c$
$\Rightarrow \log (\frac{y^2}{x^2}+1)+{\tan }^{-1}\frac{y}{x}=c+2\log x$
For $y\left(1\right)=1$
$\Rightarrow \log (\frac{1^2}{1^2}+1)+{\tan }^{-1}\frac{1}{1}=c+2\log 1$
$\Rightarrow \log 2+\frac{\pi }{2}=c$
$\therefore {\tan }^{-1}\frac{y}{x}+\log \left(\frac{x^2+y^2}{2}\right)=\frac{\pi }{4}$