Question

The solution of differential equation $(1+y^2)dx=({\tan }^{-1}y-x)dy$ is
(where $C$ is integration constant)

• A. $y={\tan }^{-1}x-1+C{e}^{-{\tan }^{-1}x}$
• B. $x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}$
• C. $y={\tan }^{-1}x+1+C{e}^{-{\tan }^{-1}x}$
• D. $x={\tan }^{-1}y+1+C{e}^{-{\tan }^{-1}y}$

Answer 1

The correct option is B $x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}$

The given equation can be written as,
$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$(Linear form)
$I.F.=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$
Solution is given by,
$x\cdot e^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}\cdot {\tan }^{-1}y}{1+y^2}dyx\cdot e^{{\tan }^{-1}y}=e^{{\tan }^{-1}y}\cdot {\tan }^{-1}y-e^{{\tan }^{-1}y}+C\because \int t{e}^{t}dt=e^t(t-1),\text{in R.H.S. take}t={\tan }^{-1}y\Rightarrow x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}$