Mathematics · 2021
JEE Main · 25 July 2021, Shift 1 · Q62
Let y=y(x) be the solution of the differential equation (d y)/(d x)=1+x e^y-x,-√ 2 <x<√ 2, y(0)=0 then, the minimum value of y(x), x ∈(-√ 2, √ 2 ) is…
Let $\displaystyle y=y(x)$ be the solution of the differential equation
$$\frac{\mathrm{d} y}{\mathrm{~d} x}=1+x \mathrm{e}^{y-x},-\sqrt{2}<x<\sqrt{2}, y(0)=0
$$
then, the minimum value of $\displaystyle y(x), x \in(-\sqrt{2}, \sqrt{2})$ is equal to :
Official answer
(1)
$\displaystyle (1-\sqrt{3})-\log _{\mathrm{e}}(\sqrt{3}-1)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.