If$y = A{e^{mx}} + B{e^{nx}}$ ,then show that
$\cfrac{{{d^2}y}}{{d{x^2}}} - (m + n)\cfrac{{dy}}{{dx}} + mny = 0$ .

Let$y = A{e^{mx}} + B{e^{nx}}$ …(i)
Differentiating (i) w.r.t. x,we get

$\cfrac{{dy}}{{dx}} = A{e^{mx}} \cdot m + B{e^{nx}} \cdot n = Am{e^{mx}} + Bn{e^{nx}}$ …(ii)

Differentiating (ii) w.r.t.x, we get

$\cfrac{{{d^2}y}}{{d{x^2}}} = Am{e^{mx}} \cdot m + Bn{e^{nx}} \cdot n = A{m^2}{e^{{\mathop{\rm m}\nolimits} x}} + B{n^2}{e^{nx}}$ …(iii)

Now, $\cfrac{{{d^2}y}}{{d{x^2}}} - (m + n)\cfrac{{dy}}{{dx}} + mny = A{m^2}{e^{{\mathop{\rm m}\nolimits} x}} + B{n^2}{e^{nx}}$

$- [(m + n)(Am{e^{mx}} + Bn{e^{nx}})] + mn(A{e^{{\mathop{\rm m}\nolimits} x}} + B{e^{nx}})$

[from (i),(ii) and (iii)]

$= A{m^2}{e^{mx}} + B{n^2}{e^{nx}} - A{m^2}{e^{mx}} - Bmn{e^{nx}} - Amn{e^{{\mathop{\rm m}\nolimits} x}} - B{n^2}{e^{nx}} + Amn{e^{mx}} + Bmn{e^{nx}} = 0$

Hence proved.