Mathematics · 2026
JEE Main · 22 January 2026, Shift 1 · Q19
Let the solution curve of the differential equation x d y-y d x=√(x^2+y^2) d x, x>0, y(1)=0, be y=y(x). Then y(3) is equal to
Let the solution curve of the differential equation $\displaystyle x d y-y d x=\sqrt{x^2+y^2} d x, x>0$, $\displaystyle y(1)=0$, be $\displaystyle y=y(x)$. Then $\displaystyle y(3)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 4$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.