Q50 / 90 · JEE Main 2024 (04 Apr Shift 1) · Mathematics

If the solution $y=y(x)$ of the differential equation $\left(x^4+2 x^3+3 x^2+2 x+2\right) \mathrm{d} y-\left(2 x^2+2 x+3\right) \mathrm{d} x=0$ satisfies $y(-1)=-\frac{\pi}{4}$, then $y(0)$ is equal to :

  • A
    $\frac{\pi}{2}$
  • B
    $-\frac{\pi}{2}$
  • C
    0
  • Correct answer
    D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & y(-1)=\frac{-\pi}{4} \\ & \frac{-\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & y(0)=\tan ^{-1} 1=\frac{\pi}{4}\end{aligned}$

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

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