Question

If $x \,d y=y d x +y^{2} d y$ and $y(1)=1$ then find the value of $y(-3)$.

Answer 1

$\int \frac{y \,d x-x\, d y}{y^{2}}=\int-d y$
$\frac{x}{y}=-y+c$
As, $y(1)=1$
$1=-1+c$
$\Rightarrow c =2$
$\Rightarrow \frac{ x }{ y }=- y +2$
If $x =-3$
$-3=- y ^{2}+2 y$
$\Rightarrow y ^{2}-2 y -3=0$
$\Rightarrow y =3$