Question

The particular integral of $\frac{d^2y}{dx^2}+2y=x^2$is

• A. $x^2 - 1$
• B. $x^2 + 1$
• C. $\frac{1}{2}(x^2-1)$
• D. $\frac{1}{2}(x^2 + 1)$

Answer 1

Answer: (C)

$\frac{1}{2}(x^2-1)$

Answer 2

$If \, \, \frac{d^2 y}{dx^3} + 2y \, =x^2$
$\Rightarrow \, \, \, (D^2 + 2)y \, = x^2 \, \, \, \, \bigg[D=\frac{d}{dx}\bigg]$
Particular integral (P.I.) $=\frac{1}{D^2 + 2} x^2$
$= \frac{1}{2\bigg(1 + \frac{D^2}{2}\bigg)} . x^2 =\frac{1}{2} \bigg(1 + \frac{D^2}{2}\bigg)^{-1} . (x^2)$
$\because \, \, (1+D)^{-1} \, = 1 - D + D^2 - D^3 + ....$
$\therefore \, \, P.I. =\frac{1}{2} . \bigg[1- \bigg(\frac{D^2}{2}\bigg)+ \bigg(\frac{D^2}{2}\bigg)^2 - ...... \bigg](x^2)$
$\Rightarrow \, \, P.I. \, = \frac{1}{2}. \bigg[x^2 \, - \frac{D^2}{2} (x^2)\bigg]$
$\Rightarrow \, \, \, P.I. \, = \frac{1}{2} . \bigg[x^2 - 1\bigg]$