Question

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

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