Question

Find general solution: $y dx+(x-y^2)dy=0$

Answer 1

y dx+(x-y2)dy = 0

⇒ydx = (y2-x)dy

\implies$$\frac {dx}{dy} = y2-$\frac xy$ = y-$\frac xy$

\implies$$\frac {dx}{dy}+$\frac xy$ = y

This is a linear differential equation of the form:

$\frac {dy}{dx}$+px = Q (where p=$\frac 1y$ and Q=y)

Now, I.F = e∫pdy = $e^{∫\frac 1y dy}$ = elog y = y

The general solution of the given differential equation is given by the relation,

x(I.F.) = ∫(Q×I.F.)dy + C

$\implies$xy = ∫(y.y)dy + C

$\implies$xy = ∫y2dy + C

$\implies$xy = $\frac {y^3}{3}$ + C

$\implies$x = $\frac {y^2}{3}$ + $\frac Cy$