Differential Equations — Class 12 Maths Solution

Given, $y = A\cos \alpha + B\sin \alpha$

$\Rightarrow$ $\frac{{dy}}{{dx}} = - \alpha A\sin \alpha x + \alpha B\cos \alpha x$

Again, differentiating both sides
w.r.t. $x$,

we get
$\frac{{{d^2}y}}{{d{x^2}}} = - A{\alpha ^2}\cos \alpha x - {\alpha ^2}B\sin \alpha x$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = - {\alpha ^2}(A\cos \alpha x + B\sin \alpha x)$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = - {\alpha ^2}y$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} + {\alpha ^2}y = 0$