Question

Solve the following differential equation :
$({\cot }^{-1}y+x)dy=(1+y^2)dx$

Answer 1

For solve the solving the given equation

$\left(1+y^2\right)\frac{dy}{dx}={\cot }^{-1}y+x$

$\Rightarrow \frac{dy}{dx}+x\left(\frac{-1}{1+y^2}\right)=\frac{{\cot }^{-1}y}{1+y^2}$

$\Rightarrow x{e}^{\int -\frac{dy}{1+y^2}}=\int e^{\int -\frac{dy}{1+y^2}}\frac{{\cot }^{-1}y}{1+y^2}dy$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=e^{\frac{-\pi }{2}}\int e^{\frac{\pi }{2}}{e}^{-{\tan }^{-1}y}\frac{{\cot }^{-1}y}{1+y^2}dy$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=e^{\frac{-\pi }{2}}\int e^{{\cot }^{-1}y}\frac{{\cot }^{-1}y}{1+y^2}dy$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=-e^{\frac{-\pi }{2}}\int e^{t}tdt$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=-e^{\frac{-\pi }{2}}\left[t{e}^t-e^t\right]$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=e^{\frac{-\pi }{2}}\left[{\cot }^{-1}y-1\right]e^{{\cot }^{-1}y}$

$\Rightarrow x{e}^{-{\tan }^{-1}y}=e^{\frac{-\pi }{2}}+C$

$x=1+C{e}^{{\tan }^{-1}y}$

Hence, which is the required answer.