Differential Equations — Class 12 Maths Solution

Given that, $y = a{e^{mx}} + b{e^{ - mx}}$

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

we get
$\frac{{dy}}{{dx}} = ma{e^{mx}} - bm{e^{ - mx}}$

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

we get
$\frac{{{d^2}y}}{{d{x^2}}} = {m^2}a{e^{mx}} + b{m^2}{e^{ - mx}}$

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} = {m^2}\left( {a{e^{mn}} + b{e^{ - mn}}} \right)$

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

$\Rightarrow$ $\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0$