Mathematics · 2019
JEE Main · 8 April 2019, Shift 1 · Q77
Let y=y(x) be the solution of the differential equation, (x^2+1)^2 (d y)/(d x)+2 x(x^2+1) y=1 such that y(0)=0. If √ a y(1)=(π)/32, then the value of…
Let $\displaystyle y=y(x)$ be the solution of the differential equation, $\displaystyle \left(x^2+1\right)^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 x\left(x^2+1\right) y=1$ such
that $\displaystyle y(0)=0$. If $\displaystyle \sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}$, then the value of ' $\displaystyle a$ ' is:
Official answer
(1)
$\displaystyle \frac{1}{16}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.