Q.$y=ae^{mx}+be^{-mx}$ satisfies which of the following differential equation?
(A) $\frac{dy}{dx}+my=0$
(B) $\frac{dy}{dx}-my=0$
(C) $\frac{d^2y}{dx^2}-m^2y=0$
(D) $\frac{d^2y}{dx^2}+m^2y=0$

Concept: Verification of Solution — substitute the given function into each differential equation to see which one holds identically.

Step 1: Differentiate $y = a e^{mx} + b e^{-mx}$.

Step 2: Differentiate again.

$\frac{d^2y}{dx^2} = a m^2 e^{mx} + b m^2 e^{-mx} = m^2 (a e^{mx} + b e^{-mx}) = m^2 y$ …

The key here is to recognise that $y = a e^{mx} + b e^{-mx}$ is the general solution of a second-order linear differential equation with constant coefficients. The characteristic roots are $m$ and $-m$, which means the auxiliary equation is $r^2 - m^2 = 0$. That directly gives the differential equation $y'' - m^2 y = 0$.

But let’s verify it step by step — not just by pattern-matching, but by actually differentiating and substituting.

1. First derivative Differentiate $y$ with respect to $x$:

Notice that this is not simply $m y$ or $-m y$, because the signs inside the bracket are different. So options (A) and (B) are not satisfied in general.

2. Second derivative Differentiate again:

But $a e^{mx} + b e^{-mx}$ is exactly $y$. So:

3. Rearrange into standard form Bring all terms to one side: $\frac{d^2y}{dx^2} - m^2 y = 0.$ …

Method: Match a given function to its differential equation by differentiating

When a function carrying arbitrary constants is given and you must decide which differential equation it satisfies, differentiate and substitute — never guess from the shape of the options.

Steps

Step 1: Count the arbitrary constants.

A function with $n$ independent constants belongs to an $n$-th order equation. Two constants (like $a$ and $b$) point to a second-order equation, so a first-order option can usually be ruled out at once.

Step 2: Differentiate as many times as there are constants.

Differentiate once, then again, watching for a derivative that reproduces the original combination up to a constant multiple. …

Common Mistakes

Mistake 1: Assuming the combination satisfies a first-order equation.

Why it's wrong: because $e^{mx}$ obeys $y'=my$, students pick option (A) or (B) for $y=ae^{mx}+be^{-mx}$. But the two exponentials have opposite-sign derivatives, so $\frac{dy}{dx}=m(ae^{mx}-be^{-mx})$ is neither $my$ nor $-my$. Correct approach: only the second derivative recombines into $m^2y$.

Mistake 2: Sign error on the second derivative. …