Question

Find general solution: $(x+3y^2) \frac {dy}{dx} = y,\ (y>0)$

Answer 1

(x+3y2)$\frac {dy}{dx}$ = y

⇒$\frac {dy}{dx}$= $\frac yx$+3y2

⇒$\frac {dx}{dy}$ = $\frac {x+3y^2}{y}$ = $\frac xy$+3y

⇒$\frac {dx}{dy}$-$\frac xy$ = 3y

This is a linear differential equation of the form:

$\frac {dx}{dy}$+px = Q (where p=-$\frac 1y$ and Q=3y)

Now, I.F. = $e^{∫pdy }$= $e^{-∫\frac 1ydy }$ = $e^{-log \ y}$ = $e^{log(\frac 1y)}$ = $\frac 1y$

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

x(I.F.) = ∫(Q×I.F.)dy+C

⇒x.$\frac 1y$ = ∫(3y.$\frac 1y$)dy+C

⇒$\frac xy$ = 3y+C

⇒x = 3y2+Cy