Question

Find the general solution of the differential equation $(x+3y^2)\frac{dy}{dx}=y(y>0)$

Answer 1

Given differential equation :

$(x+3y^2)\frac{dy}{dx}=y$

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

$\frac{dx}{dy}-\frac{x}{y}=3y$

Given differential equation is of the form
$\frac{dx}{dy}+Px=Q$

By comparing both the equations, we get

$P=\frac{-1}{y}$ and $Q=3y$

The general solution of the given differential equation is

$x(I.F)=\int (Q\times I.F.)dy+c$ ....(i)

Firstly, we need to find I.F.

$I.F.=e^{\int pdx}$

$I.F.=e^{\int \frac{-1}{y}dy}$

$I.F.=e^{-\text{log}y}$

$I.F.=e^{\text{log}{y}^{-1}}$

$I.F.=\frac{1}{y}(\because e^{\text{log}y}=y)$

Substituting the value of I.F. and $Q$ in (i),

$x\times \frac{1}{y}=\int 3y\times \frac{1}{y}.dy$

$\frac{x}{y}=\int 3dy+c$

$\frac{x}{y}=3y+c$

$x=3y^2+cy$

Hence, the required general solution is$x=3y^2+cy$