Question

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

Answer 1

Given $y=A{e}^{mx}+B{e}^{nx}$

On differentiating w.r. t. $x$, we get

$\frac{dy}{dx}=Am{e}^{mx}+Bn{e}^{nx}$

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

$\frac{d^{2}y}{d{x}^2}=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}$

We need to prove $\frac{d^{2}y}{d{x}^2}-(m+n)\frac{dy}{dx}+mny=0$

Putting them in the equation, we get

$A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-(m+n)$$(Am{e}^{mx}+Bn{e}^{nx})+mn(A{e}^{mx}+B{e}^{nx})$

$=$ $A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-$$A{m}^2{e}^{mx}-Bmn{e}^{nx}-Amn{e}^{mx}-B{n}^2{e}^{nx}+mnA{e}^{mx}+mnB{e}^{nx}$

$=0$