Question

If $y=y\left(x\right)$ is the the solution of differential equation $x^2(xdx+ydy)+2y(ydx-xdy)=0,y\left(1\right)=1$ then which of the following is/are correct ?

• A. $y\left(0\right)=0$
• B. $y\left(0\right)=3$
• C. $y\left(3\right)=3$
• D. $y\left(3\right)=9$

Answer 1

Answer: (A), (C)

$y\left(3\right)=3$

Given : $x^2(xdx+ydy)+2y(ydx-xdy)=0$
Put $x=r\cos \theta ,y=r\sin \theta$
$\Rightarrow x^2+y^2=r^2,xdx+ydy=rdr$ and $xdy-ydx=r^{2}d\theta$
the equation becomes,
$r^2{\cos }^2\theta \left(rdr\right)-2r\sin \theta \left(r^{2}d\theta \right)=0dr-2\sec \theta \tan \theta d\theta =0$
integrating both sides, we get
$r-2\sec \theta =c\Rightarrow \sqrt{x^2+y^2}-\frac{2\sqrt{x^2+y^2}}{x}=c\Rightarrow (x^2+y^2)(x-2{)}^2=C{x}^2;(C=c^2)\Rightarrow y=\sqrt{\frac{C{x}^2}{(x-2{)}^2}-x^2}$
$\Rightarrow y\left(0\right)=0$
and $y\left(1\right)=1$ (given)
$\Rightarrow C=2$
So, $y\left(3\right)=\sqrt{18-9}=3$