Q51 / 155 · WBJEE 2022 · Mathematics

If $y=e^{\tan ^{-1} x}$ then

  • Correct answer
    A
    $\left(1+x^2\right) y_2+(2 x-1) y_1=0$
  • B
    $\left(1+x^2\right) y_2+2 x y=0$
  • C
    $\left(1+x^2\right) y_2-y_1=0$
  • D
    $\left(1+x^2\right) y_2+3 x y_1+4 y=0$

Solution

$\frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{1+x^2}$ $\Rightarrow\left(1+x^2\right) y^{\prime}=y$ or $2 x y^{\prime}+\left(1+x^2\right) y^{\prime \prime}=y^{\prime}$ or $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x-1)=0$

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

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