Mathematics · 2019
JEE Main · 9 April 2019, Shift 1 · Q77
The solution of the differential equation x (d y)/(d x)+2 y=x^2(x ≠ 0) with y(1)=1, is:
The solution of the differential equation
$\displaystyle x \frac{\mathrm{~d} y}{\mathrm{~d} x}+2 y=x^2(x \neq 0)$ with $\displaystyle y(1)=1$, is :
Official answer
(3)
$\displaystyle y=\frac{x^2}{4}+\frac{3}{4 x^2}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.