Question

Which of the following differential equations has y = x as one of its particular solution?
(a) $\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=x$

(b) $\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=x$

(c) $\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0$

(d) $\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+xy=0$

Answer 1

$\left(\mathrm{c}\right)\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0$

We have,
y = x .....(1)
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{dy}{dx}=1.....\left(2\right) \\ \mathrm{Differentiating}\mathrm{again}\mathrm{with}\mathrm{respect}\mathrm{to}\mathit{}x,\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=0 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}+x^2=x^2 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}+x\times x=x^2\times 1 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}+xy=x^2\times 1\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}+xy=x^2\frac{dy}{dx}\left[\mathrm{Using}\left(2\right)\right] \\ \Rightarrow \frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0 \end{gathered}$$