Question

The solution of the differential equation $y^{2}dy=x(ydx-xdy)\text{is}y=y\left(x\right)$. If $y\left(\sqrt{3}e\right)=e$ and $y\left(x_0\right)=1$ then $x_0$ is

• A. $0$
• B. $1$
• C. $\frac{1}{e}$
• D. $-1$

Answer 1

Answer: (B)

$1$

$y^{2}dy=x(ydx-xdy)$

$1=\frac{x}{y}\frac{dy}{dx}-\frac{x^2}{y^2}.....\left(1\right)$

let $x=vy$

then, $\frac{dx}{dy}=v+y\frac{dv}{dy}$

now putting this in equation (1), we get

$1+v^2=v(v+vy\frac{dv}{dy})$

$1=vy\frac{dv}{dy}$

integrating the above, we get

$\int \frac{dy}{y}=\int vdv=logy=\frac{v^2}{2}+C$

$logy=\frac{x^2}{{2y}^2}+C$

Using the initial conditions, we get

$1=\frac{3}{2}+C⟹C=\frac{-1}{2}$

$log\left(y\right)=\frac{x^2}{{2y}^2}-\frac{1}{2}$ when $y=0,$ then

$log\left(1\right)=\frac{x^2}{{2\left(1\right)}^2}-\frac{1}{2}$

$0=\frac{x^2}{2}-\frac{1}{2}⟹\frac{x^2}{2}=\frac{1}{2}$

$x^2=1⟹x_0=1$ (Answer)

So, option (B) is correct.