Question

If $ydx+y^{2}dy=xdy$, $x\in R$, $y>0$ and $y(1)=1$, then $y(-3)=$

• A. $3$
• B. $2$
• C. $1$
• D. $5$

Answer 1

Answer: (A)

We have, $\frac{ydx-xdy}{y^2}+dy=0$
$\Rightarrow d\left(\frac{x}{y}+y\right)=0$
On integration, we get $\frac{x}{y}+y=c$
$x=1$,
$y=1$
$\Rightarrow c=2$
$\therefore \frac{x}{y}+y=2$
when $x=-3$
$\Rightarrow -\frac{3}{y}+y=2$
$\Rightarrow y^2-2y-3=0$
$\Rightarrow y=3$, $-1$