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

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> [!TLDR]
> Differentiating twice and substituting, the coefficients of $Ae^{mx}$ and $Be^{nx}$ each reduce to $0$, so the equation holds.

Given $y=Ae^{mx}+Be^{nx}$.

**First derivative:**
$$\frac{dy}{dx}=Am\,e^{mx}+Bn\,e^{nx}.$$

**Second derivative:**
$$\frac{d^2y}{dx^2}=Am^2 e^{mx}+Bn^2 e^{nx}.$$

**Substitute into the LHS:**
$$\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny$$
$$=\big(Am^2 e^{mx}+Bn^2 e^{nx}\big)-(m+n)\big(Am\,e^{mx}+Bn\,e^{nx}\big)+mn\big(Ae^{mx}+Be^{nx}\big).$$

**Collect the coefficient of $Ae^{mx}$:**
$$m^2-(m+n)m+mn = m^2-m^2-mn+mn = 0.$$ …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

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

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

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