Question
Find the general solution of $\frac{d y}{d x}+\frac{y}{x}=x^{2}$

Answer

It is given that $\frac{d y}{d x}+\frac{y}{x}=x^{2}$ 
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, p = $\frac{1}{x}$ and $Q = x^2)$
Now, $I.F. = \mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \frac{1}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}}=\mathrm{x}$ 
Thus, the solution of the given differential equation is given by the relation:
$y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{y}(\mathrm{x})=\int\left(\mathrm{x}^{2} \cdot \mathrm{x}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow x y=\int\left(x^{3}\right) d x+C$ 
$\Rightarrow x y=\frac{x^{4}}{4}+C$ 
Therefore, the required general solution of the given differential equation is $x y=\frac{x^{4}}{4}+C$

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