Mathematics · 2020
JEE Main · 9 January 2020, Shift 1 · Q74
If for x ≥ 0, y=y(x) is the solution of the differential equation, (x+1) d y=((x+1)^2+y-3) d x, y(2)=0, then y(3) is equal to \ \ \ \.
If for $\displaystyle x \geqslant 0, y=y(x)$ is the solution of the differential equation,
$$(x+1) \mathrm{d} y=\left((x+1)^2+y-3\right) \mathrm{d} x, y(2)=0,
$$
then $\displaystyle y(3)$ is equal to \_\_\_\_.
Official answer
3
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.