Question

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

Answer 1

Given that, $(x+2y^3)\frac{dy}{dx}=y$
$\Rightarrow y.\frac{dy}{dx}=x+2y^3\Rightarrow \frac{dx}{dy}=\frac{x}{y}+2y^2\Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y^2$
which is a linear differential equation.
On comparing it with $\frac{dx}{dy}+Px=Q,$ we get
$P=-\frac{1}{y},Q=2y^{2}IF=e^{\int -\frac{1}{y}dy}=e^{-\int \frac{1}{y}}dy=e^{-logy}=\frac{1}{y}$
The general solution is $x.\frac{1}{y}=\int 2y^2.\frac{1}{y}dy+C\Rightarrow \frac{x}{y}=\frac{2y^2}{2}+C\Rightarrow \frac{x}{y}=y^2+C\Rightarrow x=y^3+Cy$