Q.If $y = 3e^{2x} + 2e^{3x}$, prove that $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$.

Concept: Verification of a solution to a linear homogeneous differential equation with constant coefficients.

We are given $y = 3e^{2x} + 2e^{3x}$.

Step 1 — First derivative

Step 2 — Second derivative

Step 3 — Substitute into the left-hand side …

Why This Approach Works

The equation $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0$ is a homogeneous linear differential equation with constant coefficients. For such equations, exponential functions of the form $e^{rx}$ are natural candidates for solutions — because differentiating an exponential simply multiplies it by the constant $r$. Here, the given $y$ is a sum of two exponentials, $e^{2x}$ and $e^{3x}$. These correspond to the roots $r = 2$ and $r = 3$ of the characteristic equation $r^2 - 5r + 6 = 0$. So the problem is essentially checking that a linear combination of these exponentials indeed satisfies the differential equation.

Instead of solving the equation from scratch, we are verifying that the given $y$ works. That means we just need to compute derivatives and substitute — no guesswork, no solving.

1. Write down the function clearly.

2. Differentiate once to get $\frac{dy}{dx}$. The derivative of $e^{2x}$ is $2e^{2x}$, and of $e^{3x}$ is $3e^{3x}$. So:

3. Differentiate again to get $\frac{d^2y}{dx^2}$. Differentiate each term of $\frac{dy}{dx}$:

4. Now form the expression $\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y$. Substitute each piece:

5. Simplify term by term. First, expand the $-5$ term:

Then expand the $+6y$ term:

Now collect all $e^{2x}$ terms: …

Method: Verifying an Exponential-Sum Function Satisfies a Linear ODE

This method applies whenever $y$ is a sum of exponential terms like $e^{rx}$ and you must show it satisfies a second-order linear equation with constant coefficients (e.g. $\dfrac{d^2y}{dx^2}-p\dfrac{dy}{dx}+qy=0$).

Steps

Step 1: Differentiate each exponential term, twice

For a term $c\,e^{rx}$, each differentiation multiplies by the constant $r$:

Apply this once to get $\dfrac{dy}{dx}$, and again (on the result) to get $\dfrac{d^2y}{dx^2}$ — keep each exponential term ($e^{rx}$ for each distinct $r$) separate.

Step 2: Substitute into the given expression and group by exponential …

Common Mistakes

Mistake 1: Forgetting to multiply by the exponent's coefficient

Why it's wrong: the derivative of $e^{2x}$ is $2e^{2x}$, not $e^{2x}$ — omitting that multiplier (or using the wrong one) throws off every subsequent derivative and the final cancellation. Correct approach: apply the chain rule explicitly every time — $\frac{d}{dx}e^{rx}=r\,e^{rx}$ — and re-derive $r$ from the exponent rather than guessing it.

Mistake 2: Combining unlike exponential terms as if they were like terms …