Q. The solution of the differential equation $x\ \frac{dy}{dx} = 2y + x^{3}e^{x}$, where $y = 0$ when $x = 1$, is
A) $y = x^{3}(e^{x} - e)$
B) $y = x^{3}(e - e^{x})$
C) $y = x^{2}(e^{x} - e)$
D) $y = x^{2}(e - e^{x})$
Explanation
(c)
Given, $\frac{dy}{dx} - \frac{2}{x}\ y = x^{2}e^{x}$
$\therefore$ $IF = e^{- \int_{}^{}{\frac{2}{x}\ dx}} = e^{- \log x^{2}} = \frac{1}{x^{2}}$
$\therefore$ Complete solution is $\frac{y}{x^{2}} = \int_{}^{}\frac{x^{2}e^{x}}{x^{2}}\ dx + c$
$\Rightarrow$ $\frac{y}{x^{2}} = e^{x} + c$
$\Rightarrow$ $y = x^{2}(e^{x} + c)$
When $y = 0,\ x = 1$, then $c = - e$
$\therefore$ $y = x^{2}(e^{x} - e)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.