Solve the following differential equation 1 x dy dx = ^ -1 x,x 0 - Vidyadip

Question

Solve the following differential equation
$\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x},\text{x}\neq0$

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Answer

$\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}=\tan^{-1}\text{x},$$\text{dy}=\text{x}\tan^{-1}\text{x dx}$
$\int\text{dy}=\int\text{x}\tan^{-1}\text{x dx}$
$\text{y}-\tan^{-1}\text{x}\int\text{x dx}-\int\Big(\frac{1}{1+\text{x}^2}\int\text{x dx}\Big)\text{dx}+\text{C}$
Using intregration by part
$\text{y}=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\int\frac{\text{x}^2}{2(1+\text{x}^2)}\text{dx}+\text{C}$
$=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{\text{x}^2}{1+\text{x}^2}\text{dx}+\text{C}$
$=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\frac{1}{2}\int\Big(1-\frac{1}{\text{x}^2+1}\Big)\text{dx}+\text{C}$
$=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\frac{1}{2}\text{x}+\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$\text{y}=\frac{1}{2}(\text{x}^2+1)\tan^{-1}\text{x}-\frac{1}{2}\text{x}+\text{C}$

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