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Q. The general solution of the differential equation $x y' +y=x^{2}, x > 0$ is

KEAMKEAM 2020

Solution:

$\frac{d y}{d x}+\frac{y}{x}=x$
$I \cdot F=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
$y x=\int x \cdot x \cdot d x$
$y x=\frac{x^{3}}{3}+C$
$y=\frac{x^{2}}{3}+\frac{c}{x}$