Question

For the primitive integral equation $y d x+y^2 d y=x d y ; x \in R, y>0, y=y(x), y(1)=1$, then $y(-3)$ is

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

Answer: (A)

$ y d x+y^2 d y=x d y$
$y >0$
Divided by dy
$y \frac{d x}{d y}+y^2=x $
$\frac{d x}{d y}-\frac{1}{y} x=-y $
$\text { I.F. }=e^{-\int \frac{d y}{y}}=e^{-l n y}=\frac{1}{y}$
$x\left(\frac{1}{y}\right)=\int-y \cdot \frac{1}{y} d y $
$\frac{x}{y}=-y+c $
$x=1, y=1 \Rightarrow c=2$
$x=2 y-y^2$
$\text { when } x=-3 $
$y^2-2 y-3=0$
$(y-3)(y+1)=0 $
$y=3 $
$y=-1(\text { rejected })$