Shamim's solution: Given, $x = B \cos(\omega t + \alpha)$. Differentiating with respect to $t$: $$ \begin{aligned} \frac{dx}{dt} &= -B\omega \sin(\omega t + \alpha) \end{aligned} $$ Differentiating again with respect to $t$: $$ \begin{aligned} \frac{d^2x}{dt^2} &= -B\omega^2 \cos(\omega t + \alpha) \end{aligned} $$ Since $x = B \cos(\omega t + \alpha)$, we can write: $$ \begin{aligned} \frac{d^2x}{dt^2} &= -\omega^2 x \\ \Rightarrow \frac{d^2x}{dt^2} + \omega^2 x &= 0 \end{aligned} $$ This satisfies the given differential equation. Thus, Shamim's solution is correct. Rimjhim's solution: Given, $x = C \tan(\omega t + \varphi)$. Differentiating with respect to $t$: $$ \begin{aligned} \frac{dx}{dt} &= C\omega \sec^2(\omega t + \varphi) \end{aligned} $$ Differentiating again with respect to $t$: $$ \begin{aligned} \frac{d^2x}{dt^2} &= C\omega \cdot 2 \sec(\omega t + \varphi) \cdot \sec(\omega t + \varphi) \tan(\omega t + \varphi) \cdot \omega \\ &= 2C\omega^2 \sec^2(\omega t + \varphi) \tan(\omega t + \varphi) \end{aligned} $$ Substituting this into the differential equation: $$ \begin{aligned} \frac{d^2x}{dt^2} + \omega^2 x &= 2C\omega^2 \sec^2(\omega t + \varphi) \tan(\omega t + \varphi) + \omega^2 C \tan(\omega t + \varphi) \end{aligned} $$ Since this expression is not equal to zero, Rimjhim's solution is incorrect.