Q.Verify that the function $y = e^{-3x}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

Concept: Verification of Solution — substitute the given function and its derivatives into the differential equation and check if the equation holds.

Step 1: Compute the first derivative.

$y = e^{-3x} \quad\Rightarrow\quad \frac{dy}{dx} = -3e^{-3x}$

Step 2: Compute the second derivative.

$\frac{d^2y}{dx^2} = 9e^{-3x}$

Step 3: Substitute into the differential equation.

$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 9e^{-3x} + (-3e^{-3x}) - 6(e^{-3x})$

Step 4: Simplify.

$9e^{-3x} - 3e^{-3x} - 6e^{-3x} = (9 - 3 - 6)e^{-3x} = 0$

Since the left-hand side equals zero for all $x$, the function satisfies the equation.

The idea behind verifying a solution to a differential equation is straightforward: if a function is claimed to be a solution, then plugging it (and its derivatives) into the equation should make the equation hold true for all $x$ in the domain. Here, we have a second-order linear differential equation with constant coefficients. The given function is an exponential, which is a natural candidate because derivatives of exponentials are themselves exponentials — making substitution clean.

Let’s work through it step by step.

1. Compute the first derivative. Given $y = e^{-3x}$, differentiate with respect to $x$:

This uses the chain rule: derivative of $e^{u}$ is $e^{u} \cdot u'$, with $u = -3x$ and $u' = -3$.

2. Compute the second derivative. Differentiate $\frac{dy}{dx}$ again:

Again, the chain rule gives the factor $-3$ each time.

3. Substitute into the differential equation. The equation is $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$. Replace each term with the expressions we found:

4. Simplify the expression. Factor out $e^{-3x}$ (which is never zero, so it’s safe):

The left-hand side simplifies exactly to zero for all $x$.

Since the substitution yields $0 = 0$ identically, the function satisfies the differential equation.

Method: Verifying an exponential solves a constant-coefficient equation

To verify $y=e^{kx}$ solves a linear constant-coefficient equation, substitute its derivatives and check the coefficients sum to zero.

Steps

Step 1: Differentiate the exponential.

For $y=e^{kx}$, each derivative brings a factor $k$: $y'=k e^{kx}$, $y''=k^2 e^{kx}$.

Step 2: Substitute into the equation.

Replace $y,y',y''$ by these expressions.

Step 3: Factor out $e^{kx}$.

Since $e^{kx}\neq0$, the equation reduces to a numerical bracket, here $(k^2+k-6)$ with $k=-3$.

Step 4: Confirm the bracket is zero.

If the constants cancel to $0$, the function is a solution. (Equivalently, $k$ is a root of the characteristic equation.)

Common Mistakes

Mistake 1: Sign error differentiating $e^{-3x}$.

Why it's wrong: $\dfrac{d}{dx}e^{-3x}=-3e^{-3x}$ and $\dfrac{d^2}{dx^2}e^{-3x}=9e^{-3x}$; a lost minus makes the coefficients fail to cancel. Correct approach: apply the chain rule, bringing a factor $-3$ each time.

Mistake 2: Confusing $-6y$ with a derivative term.

Why it's wrong: $-6y=-6e^{-3x}$ uses the original function, not $y'$ or $y''$. Correct approach: substitute $y=e^{-3x}$ itself into the $-6y$ term, giving $9-3-6=0$.