Solve the differential equation (y+3x^2) dx dy =x

Question

Solve the differential equation $(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$

✓

Answer

We have,
$(\text{y}+3\text{x}^2)\frac{\text{dx}}{\text{dy}}=\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dy}}=\frac{\text{y}+3{\text{x}^{\text{2}}}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}=3{\text{x}}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=3\text{x}$
$\therefore \ \text{I}.\text{F}. = \text{e}^{\int{\text{P}\text{dx}}}$
$ =\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}{\text{y}}\Big)=\frac{1}{\text{x}}3\text{x}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^{2}}\text{y}=3$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=3\int\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$
Hence, $\frac{\text{y}}{\text{x}}=3\text{x}+\text{C}$ is the required solution.

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