Question

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

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

Answer 1

Answer: (B)

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