Q.If $y = 500e^{7x} + 600e^{-7x}$, show that $\frac{d^2 y}{dx^2} = 49y$.

Concept: Homogeneous Linear ODE with constant coefficients — the second derivative reproduces the original function scaled by a constant.

Step 1: Differentiate once.

Step 2: Differentiate again. …

We are given $y = 500e^{7x} + 600e^{-7x}$ and need to verify that $\frac{d^2 y}{dx^2} = 49y$.

The core idea here is that exponential functions of the form $e^{kx}$ have a beautiful property: differentiating them simply multiplies by $k$. So the first derivative brings down the exponent coefficient, and the second derivative brings it down twice — giving $k^2$ times the original exponential. Since both terms in $y$ have $k = 7$ or $k = -7$, and $(\pm 7)^2 = 49$, the second derivative of each term is 49 times itself. Adding them up gives exactly 49 times the original sum.

Let’s do it step by step.

1. First derivative Differentiate term by term. For $500e^{7x}$, the derivative is $500 \cdot 7 e^{7x} = 3500 e^{7x}$. For $600e^{-7x}$, the derivative is $600 \cdot (-7) e^{-7x} = -4200 e^{-7x}$. So

2. Second derivative Differentiate $\frac{dy}{dx}$ term by term. The derivative of $3500 e^{7x}$ is $3500 \cdot 7 e^{7x} = 24500 e^{7x}$. The derivative of $-4200 e^{-7x}$ is $-4200 \cdot (-7) e^{-7x} = 29400 e^{-7x}$. So

3. Factor out 49 …

Method: Verifying $y_2=k^2y$ for a Combination of $e^{kx}$ and $e^{-kx}$

Functions of the form $y=Ae^{kx}+Be^{-kx}$ have a special property: differentiating twice always returns $k^2$ times the original function, because each exponential's exponent-coefficient gets squared by two rounds of the chain rule.

Steps

Step 1: Differentiate once

$\frac{d}{dx}e^{kx}=ke^{kx}$ and $\frac{d}{dx}e^{-kx}=-ke^{-kx}$ (the chain rule brings down the exponent's own coefficient, sign included).

Step 2: Differentiate again

Each term picks up another factor of $\pm k$, and since $(\pm k)^2=k^2$ for both, the second derivative reproduces the original coefficients scaled by $k^2$. …

Common Mistakes

Mistake 1: Forgetting the chain-rule sign when differentiating $e^{-7x}$ twice.

Why it's wrong: a student who thinks "differentiating twice flips the sign each time" might expect $y_2=-49y$ for the $e^{-7x}$ term — but $(-7)^2=49$ is positive, so BOTH terms end up with the same positive $49$ coefficient. Correct approach: compute each derivative pass explicitly rather than reasoning about sign flips abstractly.

Mistake 2: Not spotting that $24500=49\times500$ and $29400=49\times600$ before declaring the result. …