Question

Find the general solution of the differential equation $ydx-(x+2y^2)dy=0$

Answer 1

The given differential equation is:

$ydx-(x+2y^2)dy=0$

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

$\Rightarrow \frac{dx}{dy}+\left(\frac{-1}{y}\right)x=2y...\left(i\right)$

This is a linear differential equation of the form

$\frac{dx}{dy}+Px=Q,\text{where}P=-\frac{1}{y}andQ=2y$

$\therefore I.F.=e^{\int Pdy}=e^{\int \frac{-1}{y}dy}=e^{\log y^{-1}}=\frac{1}{y}$
So, required solution will be:

$x.(I.F)$ = $\int Q(I.F)dyx.\frac{1}{y}=\int 2dy+C$

$(\text{using}:x(I.F.)=\int Q(I.F.)dy+C)$

$\Rightarrow \frac{x}{y}=2y+C$

which is the required solution.