Solve the following differential equation : (cot–1y + x) dy = (1 … - Vidyadip

Question

Solve the following differential equation :
(cot^–1y + x) dy = (1 + y²) dx

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Answer

$\frac{\text{dx}}{\text{dy}}-\frac{\text{x}}{1+\text{y}^2}=\frac{\cot^{-1}}{1+\text{y}^{2}}$
$\text{I.F.}=\text{e}^{-\int\frac{\text{x}}{1+\text{y}^2}}=\text{e}^{\cot^{-1}\text{y}}$
$\text{x}.\text{e}^{\cot^{-1}\text{y}}=\int\frac{\cot^{-1}\text{y}\ \text{e}^{\cot^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$
Integrating, we get
$\text{x}.\text{e}^{\cot^{-1}\text{y}}=\int\frac{\cot^{-1}\text{y}\ \text{e}^{\cot^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$
put cot^–1 y = t
$=-\int\text{t }\text{e}^{\text{t}}\text{dt}$
= (1 – t) e^t + c
⇒ x = (1 – cot^–1y) + ce^{–cot–1 y}

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