Question
For each of the differential equations given in find the general solution:
$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}(\text{y}>0)$

Answer

$(\text{x}+3\text{y}^2)\frac{\text{dy}}{\text{dx}}=\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}+3\text{y}^2}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{x}+3\text{y}^2}{\text{y}}=\frac{\text{x}}{\text{y}}+3\text{y}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}-\frac{\text{x}}{\text{y}}=3\text{y}$
This is a linear differential equation of the form:
$\frac{​​\text{dx}}{\text{dy}}+\text{px}=\text{Q}\ (\text{where p}=-\frac{1}{\text{y}}\ \text{and}\ \text{Q}=3\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{-\int\frac{\text{dy}}{\text{y}}}=\text{e}^{-\log\text{y}}=\text{e}^{\log\Big(\frac{1}{\text{y}}\Big)}=\frac{1}{\text{y}}.$
The general solution of the given differential equation is given by the relation,
$\text{x}\text{(I.F.)}=\int(\text{Q}\times\text{I.F.})\text{dy}+\text{C}$
$\Rightarrow\text{x}\times\frac{1}{​​\text{y}}=\int\Big(3\text{y}\times\frac{1}{​​\text{y}}\Big)​​\text{dy}+​​\text{C}$
$\Rightarrow\frac{\text{x}}{\text{y}}=3\text{y}+\text{C}$
$\Rightarrow\text{x}=3\text{y}^2+\text{Cy}$

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