Q72 / 180 · COMEDK 2023 (Evening Shift) · Mathematics

The general solution of the differential equation $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$

  • A
    $x=c \tan ^{-1} y+e^{-\tan ^{-1} y}$
  • B
    $x=\tan ^{-1} y+c e^{\tan ^{-1} y}$
  • Correct answer
    C
    $x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y}$
  • D
    $x=\tan ^{-1} y-1+c e^{\tan ^{-1} y}$

Solution

The given differential equation is $(1+y^2) dx = (\tan^{-1} y - x) dy$.

Rearranging the terms, we get $\dfrac{dx}{dy} = \dfrac{\tan^{-1} y - x}{1+y^2}$.

This can be written as $\dfrac{dx}{dy} + \dfrac{x}{1+y^2} = \dfrac{\tan^{-1} y}{1+y^2}$.

This is a linear differential equation of the form $\dfrac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \dfrac{1}{1+y^2}$ and $Q(y) = \dfrac{\tan^{-1} y}{1+y^2}$.

The integrating factor (IF) is $e^{\int P(y) dy} = e^{\int \dfrac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.

The solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + c$.

$x e^{\tan^{-1} y} = \int \dfrac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + c$.

Let $u = \tan^{-1} y$, then $du = \dfrac{1}{1+y^2} dy$.

The integral becomes $\int u e^u du = u e^u - \int e^u du = u e^u - e^u = e^u(u-1)$.

Substituting back, $x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + c$.

Dividing by $e^{\tan^{-1} y}$, we get $x = \tan^{-1} y - 1 + c e^{-\tan^{-1} y}$.

Answer: $x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y}$

Hence, the correct answer is option: 'C'.

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