Question
Solve the following differential equation : $\left(y+3 x^2\right) \frac{d x}{d y}=x$

Answer

From the given differential equation,
$\left(y+3 x^2\right) \frac{d x}{d y} =x$
 or $\frac{d x}{d y} =\frac{x}{y+3 x^2}$
or  $\frac{d y}{d x} =\frac{y+3 x^2}{x}$
$\Rightarrow \frac{d y}{d x} =\frac{y}{x}+3 x$
$\Rightarrow \frac{d y}{d x}+\left(-\frac{y}{x}\right) =3 x$
Comparing equation $(1)$ with linear differential equation
$\frac{d y}{d x}+P y=Q,$
Here $P =-\frac{1}{x}$, and $Q =3 x$
$\therefore$ Integrating factor
$\text { I.F. } =e^{\int Pd \ d x}$
$=e^{-\int \frac{1}{x} d x}=e^{-\log x}$
$=e^{\log (x)^{-1}}=(x)^{-1}$
$ =\frac{1}{x}$
Hence the required solution will be :
$y . \text { I.F. } =\int((I . F . Q) d x+C$
$\Rightarrow y \times \frac{1}{x} =\int \frac{1}{x} \times 3 x\ d x+C$
$\Rightarrow \frac{y}{x}=\int 3 d x+C$
$\Rightarrow \frac{y}{x}=3 x+C $
$\therefore y=3 x^2+C x$
which is the general solution of the given differential equation.

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