Solve the following differential equations:(1+y^2) ^ -1 xdx+2y(1+… - Vidyadip

Question

Solve the following differential equations:$(1+\text{y}^2)\tan^{-1}\text{xdx}+2\text{y}(1+\text{x}^2)\text{dy}=0$

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Answer

$(1+\text{y}^2)\tan^{-1}\text{xdx}+2\text{y}(1+\text{x}^2)\text{dy}=0$
$(1+\text{y}^2)\tan^{-1}\text{xdx}=-2\text{y}(1+\text{x}^2)\text{dy}$
$-\frac{\tan^{-1}}{2(1+\text{x}^2)}\text{dx}=\frac{\text{y}}{(1+\text{y}^2)}\text{dy}$
Integrating on both the sides
$\int-\frac{\tan^{-1}\text{x}}{2(1+\text{x}^2)}\text{dx}=\int\frac{\text{y}}{(1+\text{y}^2)}\text{dy}$
$-\Big(\tan^{-1}\text{x}\Big(\frac{1}{2}\tan^{-1}\text{x}\Big)-\int\frac{1}{(1+\text{x}^2)}\Big(\frac{1}{2}\tan^{-1}\text{x}\Big)\text{dx}\Big)=\frac{1}{2}\ln(\text{y}^2+1)+\text{C}$
$-\frac{1}{4}(\tan^{-1}\text{x})^2=\frac{1}{2}\ln(\text{y}^2+1)+\text{C}_1$
$\frac{1}{2}(\tan^{-1}\text{x})^2+\ln(\text{y}^2+1)=\text{C}$

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