Question

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

• A. 3
• B. 2
• C. 1
• D. 5

Answer 1

Answer: (A)

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