Solve the following differential equation:(x^ 2 +1) dy dx +2xy= x… - Vidyadip

Question

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

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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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