Q71 / 120 · VITEEE 2006 · Mathematics
The particular integral of $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \mathrm{y}=\mathrm{x}^{2}$ is
- A$\mathrm{x}^{2}-1$
- B$x^{2}+1$
- Correct answerC$\frac{1}{2}\left(\mathrm{x}^{2}-1\right)$
- D$\frac{1}{2}\left(x^{2}+1\right)$
Solution
If $\frac{d^{2} y}{d x^{2}}+2 y=x^{2}$
$\Rightarrow\left(D^{2}+2\right) y=x^{2} \quad\left[D \equiv \frac{d}{d x}\right]$
Particular integral $($ P.I. $)=\frac{1}{\mathrm{D}^{2}+2} \cdot \mathrm{x}^{2}$
$=\frac{1}{2\left(1+\frac{\mathrm{D}^{2}}{2}\right)} \cdot \mathrm{x}^{2}=\frac{1}{2} \cdot\left(1+\frac{\mathrm{D}^{2}}{2}\right)^{-1} \cdot\left(\mathrm{x}^{2}\right)$
$\because(1+\mathrm{D})^{-1}=1-\mathrm{D}+\mathrm{D}^{2}-\mathrm{D}^{3}+\ldots \ldots$
$\therefore$ P.I $=\frac{1}{2} \cdot\left[1-\left(\frac{\mathrm{D}^{2}}{2}\right)+\left(\frac{\mathrm{D}^{2}}{2}\right)^{2}-\ldots .\right]\left(\mathrm{x}^{2}\right)$
$\Rightarrow$ P.I. $=\frac{1}{2} \cdot\left[\mathrm{x}^{2}-\frac{\mathrm{D}^{2}}{2}\left(\mathrm{x}^{2}\right)\right]$
$\Rightarrow$ P.I. $=\frac{1}{2} \cdot\left[\mathrm{x}^{2}-1\right]$
Hence, the correct answer is option: 'C'.