Question

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

Hint:

Differentiate twice and substitute; the coefficients of e^{mx} and e^{nx} each vanish identically.

Answer 1

Step 1: Recognising the characteristic equation viewpoint:
\(y=Ae^{mx}+Be^{nx}\) is exactly the general solution of a 2nd-order linear ODE whose characteristic equation has roots \(m\) and \(n\), i.e. \((D-m)(D-n)y=0\), which expands to \(D^2-(m+n)D+mn\).

Step 2: Expanding the operator form:
\((D-m)(D-n)=D^2-nD-mD+mn=D^2-(m+n)D+mn\), exactly matching the required equation's coefficients.

Step 3: Confirming by direct substitution:
Since \(D e^{mx}=me^{mx}\) and \(D^2e^{mx}=m^2e^{mx}\) (similarly for \(n\)), each exponential term individually satisfies \((D-m)(D-n)\big(\cdot\big)=0\) when the matching root is used, so the linear combination does too.

Final Answer:
\[ \boxed{\dfrac{d^2y}{dx^2}-(m+n)\dfrac{dy}{dx}+mny=0} \]