Question

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

• A. $x-2 = Ke^{-\tan^{-1}}y$
• B. $2xe^{\tan^{-1}}y = e^2{^{\tan^{-1}}}y+K$
• C. $xe^{\tan^{-1}}y$ = ${{\tan^{-1}}}y+K$
• D. $x^2e^{2\tan^{-1}}y$ = $e{^{\tan^{-1}}}y+K$

Answer 1

Answer: (B)

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