Question

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

• 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. $x^2{e}^{2{\tan }^{-1}}y$ = $e{}^{{\tan }^{-1}}y+K$

Answer 1

Answer: (B)

$\left(1+y^2\right)+\left(x-e^{ta{n}^{-1}y}\right)\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}+\frac{x}{1+y^2}=\frac{e^{ta{n}^{-1}y}}{1+y^2}$
This is linear in $x$. [Type $\frac{dy}{dx}+Px=Q$]
$\int pdy=\int \frac{dy}{1+y^2}=ta{n}^{-1}y$
$\therefore e^{\int pdy}=e^{ta{n}^{-1}y}$
$\therefore$ Sol. is
$x.e^{ta{n}^{-1}y}=\int e^{ta{n}^{-1}y}\cdot \frac{e^{ta{n}^{-1}y}}{1+y^2}dy+C$
Put $ta{n}^{-1}y=z$
$\therefore \frac{dy}{1+y^2}=dz$
$\therefore x{e}^{ta{n}^{-1}y}=\int e^{2z}dz+C=\frac{1}{2}{e}^{2z}+C$
$=\frac{1}{2}{e}^{2ta{n}^{-1}y}+K$ [where $2C=K$]