Question

The solution of primitive integral equation $(x^2+y^2)dy=xy.dx$ is $y=y\left(x\right)$. If $y\left(1\right)=1$ and $y\left(x_0\right)=e$ then $x_0$ is

• A. $\sqrt{2(e^2-1)}$
• B. $\sqrt{2(e^2+1)}$
• C. $\sqrt{3}e$
• D. $\sqrt{\frac{1}{2}(e^2+1)}$

Answer 1

Answer: (C)

$\sqrt{3}e$

We have, $(x^2+y^2)dy=xydx$

$\Rightarrow \frac{dy}{dx}=\frac{xy}{x^2+y^2}$

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

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

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

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

Integrating both sides we get, $-\frac{1}{2v^2}+\log v=-\log x+\log c$, where $c$ is constant

$\Rightarrow -\frac{1}{2v^2}+\log \left(\frac{vx}{c}\right)=0$

$\Rightarrow y=c{e}^{x^2/2y^2}$

Now using $y\left(1\right)=1\Rightarrow c=e^{-1/2}$

Also $y\left(x_0\right)=e\Rightarrow e=e^{-1/2+x_0^2/2e^2}\Rightarrow 1=-1/2+x_0^2/2e^2$

$\Rightarrow x_0^2=3e^2\Rightarrow x_0=\sqrt{3}e$