Question

For the differential equation $ydx+y^{2}dy=xdy,x\in R,y\left(1\right)=1,$ then the value of $y(-3)$ is:

• A. $3$
• B. $4$
• C. $-1$
• D. $-2$

Answer 1

Answer: (A), (C)

$-1$

The given equation is $ydx+y^{2}dy=xdy;x\in R,y\left(1\right)=1$
$\Rightarrow \frac{ydx-xdy}{y^2}+dy=0$
$\Rightarrow d\left(\frac{x}{y}\right)+dy=0$
On integrating, we get $\frac{x}{y}+y=C$
$y\left(1\right)=1\Rightarrow 1+1=C\Rightarrow C=2$
$\therefore \frac{x}{y}+y=2$
Now to find $y(-3),$ putting $x=-3$ in the above equation
we get, $-\frac{3}{y}+y=2$
$\Rightarrow y^2-2y-3=0\Rightarrow y=3,-1$