Question

The solution of the differential equation $(x^2+y^2)dy=xydx$ 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. $\sqrt{\frac{e^2+1}{2}}$

Answer 1

The correct option is C $\sqrt{3}e$

The given differential equation is $(x^2+y^2)dy=xydx$ such that $y\left(1\right)=1$ and $y\left(x_0\right)=e$
The given equation can be written as:
$\frac{dy}{dx}=\frac{xy}{x^2+y^2},$ which is a homogeneous equation.
Put $y=vx$ to get $v+x\frac{dv}{dx}=\frac{v}{1+v^2}$
$\Rightarrow x\frac{dv}{dx}=\frac{-v^3}{1+v^2}$
$\Rightarrow \int \frac{1+v^2}{v^3}dv+\int \frac{dx}{x}=0$
$\Rightarrow -\frac{1}{2v^2}+\ln |v|+\ln |x|=C$
$\Rightarrow \ln |y|=C+\frac{x^2}{2y^2}$
Also $y\left(1\right)=1$
$\Rightarrow \ln 1=C+\frac{1}{2}\Rightarrow C=-\frac{1}{2}$
$\therefore \ln |y|=\frac{x^2-y^2}{2y^2}$
Given $y\left(x_0\right)=e$
$\Rightarrow \ln |e|=\frac{x_0^2-e^2}{2e^2}$
$\Rightarrow x_0^2=3e^2\Rightarrow x_0=\sqrt{3}e$