Question

If \(y=Ae^{mx}+Be^{nx}\), then show that \(\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0\).

Hint:

Differentiate twice, substitute into the given expression, and show the coefficients of e^mx and e^nx each vanish.

Answer 1

Step 1: Recognize m and n as roots of an auxiliary quadratic:
The claimed equation $y''-(m+n)y'+mny=0$ has auxiliary equation $t^2-(m+n)t+mn=0$, which factors as $(t-m)(t-n)=0$ — its roots are exactly $m$ and $n$.

Step 2: Recall that the general solution of a linear ODE with auxiliary roots $m,n$ is $y=Ae^{mx}+Be^{nx}$:
This is precisely the given form of $y$, so it must satisfy that ODE by construction of the theory.

Step 3: Confirm directly by substitution (short check):
For the $e^{mx}$ term alone: $m^2-(m+n)m+mn=(m-m)(m-n)=0$ since $t=m$ is a root of $(t-m)(t-n)$. Same for the $e^{nx}$ term with $t=n$.

Final Answer:
Both exponential pieces individually satisfy the ODE (since $m,n$ are its characteristic roots), so their linear combination does too. \[ \boxed{y''-(m+n)y'+mny=0} \]