Question

For the given differential equation find the general solution.

$(x+3y^2)\frac{dy}{dx}=y(y>0).$

Answer 1

Given, $(x+3y^2)\frac{dy}{dx}=y$ $\Rightarrow y\frac{dx}{dy}=x+3y^2$
On dividing by y both sides, we get
$\frac{dx}{dy}=\frac{x}{y}+3y\Rightarrow \frac{dx}{dy}-\frac{x}{y}=3y$
On comparing with the form $\frac{dx}{dy}+Px=Q,$ we get
$P=-\frac{1}{y},Q=3y\therefore IF=e^{-\int \frac{1}{y}dy}\Rightarrow IF=e^{-logy}=e^{log(y{)}^{-1}}=y^{-1}=\frac{1}{y}$ ...(i)
The general solution of the given differential equations is given by
$x.IF=\int Q\times IFdy+C\Rightarrow x.\frac{1}{y}=\int \frac{1}{y}\times 3ydy+C\Rightarrow \frac{x}{y}=3y+C\Rightarrow x=3y^2+Cy$