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

Question

Accepted Answer

> [!TLDR]
> The given function $y = Ae^{mx} + Be^{nx}$ is a linear combination of two exponentials. Differentiating twice and substituting into the differential equation shows that the expression simplifies to zero because the exponentials are eigenfunctions of the derivative operator, and the coefficients $(m+n)$ and $mn$ are chosen to cancel the terms.

## Why This Works: The Concept

This problem is about **linear differential equations with constant coefficients**. When you have a function like $y = Ae^{mx} + Be^{nx}$, each exponential term $e^{mx}$ has a special property: differentiating it just multiplies it by $m$. So $e^{mx}$ is an *eigenfunction* of the derivative operator.

The differential equation $\frac{d^2 y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0$ is designed so that when you plug in $e^{mx}$ or $e^{nx}$, the result is zero. Think of it as a "characteristic equation" $(r - m)(r - n) = 0$ in disguise — the numbers $m$ and $n$ are precisely the roots of $r^2 - (m+n)r + mn = 0$.

Since the differential equation is linear (no products of $y$ with its derivatives), if each exponential satisfies it individually, then any linear combination $Ae^{mx} + Be^{nx}$ also satisfies it. That's the core idea.

## Step-by-Step Verification

**1. Write down the given function and find the first derivative.**

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

Differentiating term by term:
- The derivative of $Ae^{mx}$ is $A \cdot m e^{mx} = Am e^{mx}$.
- The derivative of $Be^{nx}$ is $B \cdot n e^{nx} = Bn e^{nx}$.

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

**2. Find the second derivative.**

Differentiate $\frac{dy}{dx}$:
- The derivative of $Am e^{mx}$ is $Am \cdot m e^{mx} = Am^2 e^{mx}$.
- The derivative of $Bn e^{nx}$ is $Bn \cdot n e^{nx} = Bn^2 e^{nx}$.

So:
$$\frac{d^2 y}{dx^2} = Am^2 e^{mx} + Bn^2 e^{nx}$$

**3. Build the expression we need to check.**

We want to verify that:
$$\frac{d^2 y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0$$

Substitute each piece:

- First term: $\fr

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 $\frac{d^2 y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0$ .

Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

The definition and notation

Seeing the pattern

Why This Works: The Concept

Step-by-Step Verification

Method: Verifying an Exponential Satisfies a Constant-Coefficient Linear ODE

Steps

Common Mistakes

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

Why This Works: The Concept

Step-by-Step Verification

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

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