Question

The solution of the differential equation $(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

• A. $(x-2)=k{e}^{-{\tan }^{-1}y}$
• B. $2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$
• C. $x{e}^{{\tan }^{-1}y}={\tan }^{-1}y+k$
• D. $x{e}^{2{\tan }^{-1}y}=e^{{\tan }^{-1}y}+k$

Answer 1

The correct option is B $2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$

Consider, $1+y^2+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

$⟹(1+y^2)dx=(e^{{\tan }^{-1}y}-x)dy$
$⟹\frac{dx}{dy}=\frac{e^{{\tan }^{-1}y}-x}{1+y^2}$
$⟹\frac{dx}{dy}+\frac{1}{1+y^2}\cdot x=\frac{e^{{\tan }^{-1}y}}{1+y^2}$
Which is the linear different equation of the form $\frac{dx}{dy}+Rx=S$, where $R$ and $S$ are functions of $y$ or constant (s)
$\therefore I.F=\int \frac{1}{1+y^2}\cdot dy=e^{{\tan }^{-1}y}$
Hence required solution is
$x.(I.F.)=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}=(I.F)dy$
$⟹x.e^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}\left(e^{{\tan }^{-1}y}\right)dy$
$⟹x.e^{{\tan }^{-1}y}=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy....\left(1\right)$
Put $t={\tan }^{-1}y$
$\therefore \frac{dt}{dy}=\frac{1}{1+y^2}\Rightarrow dt=\frac{1}{1+y^2}.dy$
$\therefore \int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy=\int e^{2t}.dt=\frac{e^{2t}}{2}+K$
Hence equation $\left(1\right)$ becomes,
$x{e}^{{\tan }^{-1}y}=\frac{1}{2}{e}^{2t}+K$
$⟹x{e}^{{\tan }^{-1}y}=\frac{1}{2}{e}^{2{\tan }^{-1}y}+K$
$⟹2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+K$ is the required solution.