Mathematics · 2022
JEE Main · 29 June 2022, Shift 1 · Q2
Let the solution curve of the differential equation x (d y)/(d x)-y=√(y^2+16 x^2), y(1)=3 be y=y(x). Then y(2) is equal to:
Let the solution curve of the differential equation
$\displaystyle x \frac{\mathrm{~d} y}{\mathrm{~d} x}-y=\sqrt{y^2+16 x^2}, y(1)=3$ be $\displaystyle y=y(x)$. Then $\displaystyle y(2)$ is equal to :
Official answer
(1)
$\displaystyle 15$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.