Question

If $\frac{dy}{dx}=\frac{x-y}{x+y}$ and $y\left(1\right)=1$ then $y\left(2\right)$ equals

• A. $5$
• B. $2\pm \sqrt{10}$
• C. $2\pm \sqrt{12}$
• D. $-2\pm \sqrt{10}$

Answer 1

The correct option is D $-2\pm \sqrt{10}$

Given $\frac{dy}{dx}=\frac{x-y}{x+y}$

Substituting $y=xv$

Differentiating w.r.t. x, we get,

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

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

$\Rightarrow y\left(x\right)=-\sqrt{c+2x^2}-x$ or $y\left(x\right)=\sqrt{c+2x^2}-x$

$y\left(1\right)=1\Rightarrow -\sqrt{c+2}-1=1\Rightarrow c=2$

Therefore,

$y\left(2\right)=-2\pm \sqrt{10}$