Question

If the solution curve of the differential equation $\left(\left({\tan }^{-1}y\right)-x\right)dy=\left(1+y^2\right)dx$ passes through the point $(1,0)$ then the abscissa of the point on the curve whose ordinate is $\tan (1)$ is :

• A. $2e$
• B. $\frac{2}{e}$
• C. $2$
• D. $\frac{1}{e}$

Answer 1

Answer: (B)

$\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}$
I.f $=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$
$x{e}^{{\tan }^{-1}y}=\int \frac{{\tan }^{-1}y}{1+y^2}{e}^{{\tan }^{-1}y}dy$
$x\cdot e^{{\tan }^{-1}y}=\left({\tan }^{-1}y-1\right)e^{{\tan }^{-1}y}+c$
$\therefore (1,0)$ lies exit $c=2$.
For $y=\tan 1\Rightarrow x=\frac{2}{e}$