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 :

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Differential Equations

If the solution curve of the differential equation

((tan^{- 1} y) - x) d y = (1 + y^{2}) d x

passes through the point

(1, 0)

then the abscissa of the point on the curve whose ordinate is

tan (1)

is :

Answer: (B)

Rewriting the differential equation:

\frac{d x}{d y} + \frac{x}{1 + y ^{2}} = \frac{t a n ^{- 1} y}{1 + y ^{2}}

This is a linear first-order ODE. The integrating factor is

e^{\int \frac{1}{1 + y ^{2}} d y} = e^{t a n^{- 1} y}

.

Multiplying through:

x e^{t a n^{- 1} y} = \int \frac{t a n ^{- 1} y}{1 + y ^{2}} e^{t a n^{- 1} y} d y

Using integration by parts with

u = tan^{- 1} y

x e^{t a n^{- 1} y} = (tan^{- 1} y - 1) e^{t a n^{- 1} y} + c

Applying the initial condition

(1, 0)

1 \cdot 1 = (0 - 1) \cdot 1 + c

, so

c = 2

.

For

y = tan (1)

, we have

tan^{- 1} y = 1

x e^{1} = (1 - 1) e^{1} + 2

x = \frac{2}{e}

Answer: (B)

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option