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

Concept: Second-order derivative of trigonometric functions.

We are given $y = 5\cos x - 3\sin x$.

Step 1: Differentiate once.

Step 2: Differentiate again to get the second derivative.

Step 3: Add $y$ to this result. …

Why This Works: The Concept

This is a classic verification problem for a homogeneous linear differential equation with constant coefficients. The equation $\frac{d^2 y}{dx^2} + y = 0$ is the simple harmonic oscillator equation — its solutions are precisely linear combinations of $\sin x$ and $\cos x$.

What makes it work is the cyclic nature of derivatives of sine and cosine:

• $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$
• So $\frac{d^2}{dx^2}(\sin x) = -\sin x$ and $\frac{d^2}{dx^2}(\cos x) = -\cos x$

That means for any constants $A$ and $B$, the function $y = A\cos x + B\sin x$ will satisfy $y'' + y = 0$ — because the second derivative just flips the sign of each term, and adding the original function cancels everything. Here $A = 5$ and $B = -3$, so the same logic applies.

Step-by-Step Verification

1. Write down the given function.

We have

2. Find the first derivative $\frac{dy}{dx}$.

Differentiate term by term:

• Derivative of $5\cos x$ is $5(-\sin x) = -5\sin x$
• Derivative of $-3\sin x$ is $-3(\cos x) = -3\cos x$

So

3. Find the second derivative $\frac{d^2 y}{dx^2}$.

Differentiate $\frac{dy}{dx}$:

• Derivative of $-5\sin x$ is $-5\cos x$
• Derivative of $-3\cos x$ is $-3(-\sin x) = 3\sin x$

Thus

Method: Verifying a Function Satisfies a Second-Order ODE

This method applies whenever you're asked to "prove" or "show" that a given $y$ satisfies a stated relation involving $y$, $y_1$, and $y_2$.

Steps

Step 1: Compute $y_1=\dfrac{dy}{dx}$

Differentiate term by term.

Step 2: Compute $y_2=\dfrac{d^2y}{dx^2}$

Differentiate $y_1$ again.

Step 3: Substitute $y_2$ (and $y_1$, $y$ if needed) into the stated expression and simplify

The goal is to show the expression collapses to the claimed value (often $0$). …

Common Mistakes

Mistake 1: Sign errors across the two differentiation passes.

Why it's wrong: $\frac{d}{dx}(5\cos x)=-5\sin x$ then $\frac{d}{dx}(-5\sin x)=-5\cos x$ — each pass flips or preserves a sign, and losing track after two steps is common, especially with the $-3\sin x$ term. Correct approach: differentiate one term completely through both passes before moving to the next term.

Mistake 2: Assuming the cyclic pattern without actually computing both derivatives. …