Question
Solve the following differential equation :
$
x d y+\left(y-x^3\right) d x=0
$

Answer

The given differential equation is
or $ \ \ \ \ \
x d y+\left(y-x^3\right) d x=0
$
or $\quad x \frac{d y}{d x}+y=x^3$
or $\quad \frac{d y}{d x}+\frac{1}{x} y=x^2$
$
x \frac{d y}{d x}+y-x^3=0
$
Comparing equation (1) with the linear differential equation $\frac{d y}{d x}+ P y= Q$
Here $P =\frac{1}{x}$ and $Q =x^2$
Integrating factor $
\text { I.F. }=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x
$
Hence the required solution of the differential equation will be :
$
\begin{aligned}
y \times \text { I.F. } & =\int(I . F) Q d x+C \\
\Rightarrow \quad y \times x & =\int(x) \cdot x^2 d x+C \\
\Rightarrow \quad y x & =\int x^3 d x+C \\
\Rightarrow \quad & =\frac{x^4}{4}+C \\
\Rightarrow \quad y & =\frac{1}{4} x^3+\frac{C}{x}
\end{aligned}
$
This is the general solution of the given differential equation.

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