Solve the following differential equation : (y+3 x^2 ) d x d y =x

Question

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

✓

Answer

From the given differential equation,
$
\begin{array}{rlrl}
& & \left(y+3 x^2\right) \frac{d x}{d y} & =x \\
\text { or } & \frac{d x}{d y} & =\frac{x}{y+3 x^2} \\
\text { 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
\end{array}
$
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. } \begin{aligned}
& =e^{\int Pd d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x)^{-1}}=(x)^{-1} \\
& =\frac{1}{x}
\end{aligned}
$
Hence the required solution will be :
$
\begin{aligned}
y . \text { I.F. } & =\int((I . F . Q) d x+C \\
\Rightarrow \quad y \times \frac{1}{x} & =\int \frac{1}{x} \times 3 x d x+C
\end{aligned}
$
$
\begin{array}{l}
\Rightarrow \quad \frac{y}{x}=\int 3 d x+C \\
\Rightarrow \quad \frac{y}{x}=3 x+C \quad \therefore y=3 x^2+C x
\end{array}
$
which is the general solution of the given differential equation.

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