Question
Solve the following differential equation:
$(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+\text{2xy}=\sqrt{\text{x}^{2}+4}$.

Answer

$(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+\text{2xy}=\sqrt{\text{x}^{2}+4}$
$\frac{\text{dy}}{\text{dx}}+\frac{\text{2x}}{\text{x}^{2}+1}\text{y}=\frac{\sqrt{\text{x}^{2}+4}}{\text{x}^{2}+1}$
$\text{I.F.}=\text{e}^{\int\frac{\text{2x}}{\text{x}^{2}+1}\text{dx}}=(\text{x}^{2}+1)$
Solution is $ y.(x^2 + 1) =\int\sqrt{\text{x}^{2}+4}\text{ dx + c}$
$y (x^2 + 1) =\frac{1}{2}\text{x}\sqrt{\text{x}^{2}+4}+2\log\Big(\text{x}+\sqrt{\text{x}^{2}+4}\Big)+\text{c.}$

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