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

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Accepted Answer

> [!TLDR]
> Differentiating $y=Ae^{mx}+Be^{nx}$ twice and substituting makes the coefficients of $e^{mx}$ and $e^{nx}$ vanish, proving the relation.

**Step 1 — First derivative.**
$$y=Ae^{mx}+Be^{nx}$$
$$\frac{dy}{dx}=Ame^{mx}+Bne^{nx}.$$

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

**Step 3 — Substitute into the left side.**
$$\frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mn\,y$$
$$=\big(Am^2e^{mx}+Bn^2e^{nx}\big)-(m+n)\big(Ame^{mx}+Bne^{nx}\big)+mn\big(Ae^{mx}+Be^{nx}\big).$$

**Step 4 — 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 $y = Ae^{mx} + Be^{nx}$ then show that $\dfrac{d^2 y}{dx^2} - (m + n)\dfrac{dy}{dx} + mn\, y = 0$ .

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Q.If y=Aemx+Benxy = Ae^{mx} + Be^{nx}y=Aemx+Benx then show that d2ydx2−(m+n)dydx+mn y=0\dfrac{d^2 y}{dx^2} - (m + n)\dfrac{dy}{dx} + mn\, y = 0dx2d2y​−(m+n)dxdy​+mny=0.

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

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