Question

For the primitive integral equation $ydx+y^{2}dy=xdy;x\in R,y>0,y\left(x\right),y\left(1\right)=-1$, then $y(-3)$ is:

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

Answer: (C)

1

$ydx+y^{2}dy=xdy$
$\Rightarrow \frac{ydx-xdy}{y^2}=-dy$
$\Rightarrow \frac{x}{y}=-y+c$
$=>y\left(1\right)=-1\Rightarrow c=-2\Rightarrow y^2+2y+x=0$
$=>y(-3):y^2+2y-3=0\Rightarrow (y+3)(y-1)=0=>y=-3,1$