Mathematics - Differential Equation Question with Solution | TestHub

Given differential equation,$ydx-xdy-3y^2 dy=0$

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

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

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

$\therefore$solution is

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

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

Passes through$(1,1)$,

$\therefore 1=3+c ;c=-2$

Equation of required curve,$x=3y^2-2y$

Clearly, it passes through$\left(\frac{-1}{3},\frac{1}{3}\right)$.