Question

If $xdy=ydx+y^{2}dy,y>0$ and y(1) = 1, then what is y(-3) equal to?

• A. 3 only
• B. -1 only
• C. Both -1 and 3
• D. Neither -1 nor 3

Answer 1

The correct option is A 3 only

Given Differential equation is $xdy=ydx+y^{2}dy$,

$⟹\frac{dx}{dy}=\frac{x}{y}-y$,

$⟹\frac{dx}{dy}-\frac{x}{y}=-y$ this is a linear differential equation in $y$,

The integrating factor is $\frac{1}{y}$,

Now the differential equation becomes,

$\frac{1}{y}\frac{dx}{dy}-\frac{x}{y^2}=-1$,

$⟹\frac{x}{y}=\int -1dy⟹\frac{x}{y}=-y+c$,

Given that $y\left(1\right)=1$,

$c=2$,

to find $y(-3)$,

$⟹y^2-2y-3=0$,

Solving the quadratic equation gives,

$y=-1andy=3$,

Given that $y>0$,

$\therefore y=3only$