jee-main 2019 Q86

jee-main · India · session2_08apr_shift1 First order differential equations (integrating factor)

Let $y = y(x)$ be the solution of the differential equation, $\left(x^2 + 1\right)^2 \frac{dy}{dx} + 2x\left(x^2 + 1\right)y = 1$ such that $y(0) = 0$. If $\sqrt{a}\, y(1) = \frac{\pi}{32}$, then the value of $a$ is
(1) $\frac{1}{16}$
(2) $\frac{1}{2}$
(3) $\frac{1}{4}$
(4) $1$

Let $y = y(x)$ be the solution of the differential equation, $\left(x^2 + 1\right)^2 \frac{dy}{dx} + 2x\left(x^2 + 1\right)y = 1$ such that $y(0) = 0$. If $\sqrt{a}\, y(1) = \frac{\pi}{32}$, then the value of $a$ is\\
(1) $\frac{1}{16}$\\
(2) $\frac{1}{2}$\\
(3) $\frac{1}{4}$\\
(4) $1$

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