Mathematics · 2019
JEE Main · 12 April 2019, Shift 1 · Q77
Consider the differential equation, y^2 d x+(x-1/y) d y=0. If value of y is 1 when x=1, then the value of x for which y=2, is:
Consider the differential equation, $\displaystyle y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If value of $\displaystyle y$ is $\displaystyle 1$ when $\displaystyle x=1$, then the value of $\displaystyle x$ for which $\displaystyle y=2$, is :
Official answer
(3)
$\displaystyle \frac{3}{2}-\frac{1}{\sqrt{e}}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.