Question

Solve $(1+y^2)dx=({\tan }^{-1}y-x)dy.$

• A. $\therefore x={\tan }^{-1}y-1-c{e}^{-{\tan }^{-1}y}.$
• B. $\therefore x=-{\tan }^{-1}y-1+c{e}^{-{\tan }^{-1}y}.$
• C. $\therefore x={\tan }^{-1}y-1+c{e}^{-{\tan }^{-1}y}.$
• D. None of these.

Answer 1

Answer: (C)

$\therefore x={\tan }^{-1}y-1+c{e}^{-{\tan }^{-1}y}.$

Given, $(1+y^2)dx=({\tan }^{-1}y-x)dy$

$\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$ ...(1)

Here $P=\frac{1}{1+y^2}$

$\Rightarrow \int PdP=\int \frac{1}{1+y^2}dy={\tan }^{-1}y$
$\therefore I.F=e^{{\tan }^{-1}y}.$

Multiplying (1) by $I.F.$ we get

$e^{{\tan }^{-1}y}\frac{dx}{dy}+e^{{\tan }^{-1}y}\frac{x}{1+y^2}=e^{{\tan }^{-1}y}\frac{{\tan }^{-1}y}{1+y^2}$

Integrating both sides, we get
$x{e}^{{\tan }^{-1}y}=\int e^{{\tan }^{-1}y}.\frac{{\tan }^{-1}y}{1+y^2}dy=\int t{e}^{t}dt$
where ${\tan }^{-1}y=t$
$\therefore \frac{1}{1+y^2}dy=dt$
$\Rightarrow x{e}^{{\tan }^{-1}y}=e^t.(t-1)+c=e^{{\tan }^{-1}y}({\tan }^{-1}y-1)+c.$
$\therefore x={\tan }^{-1}y-1+c{e}^{-{\tan }^{-1}y}.$