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Worked Examples · Example 36

Q.If y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x, then prove that d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0.

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✓ Free question

For a function of the form y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x, its second derivative is −y-y, so d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0 holds identically — this is a direct consequence of the fact that sine and cosine are eigenfunctions of the second derivative operator with eigenvalue −1-1.

Why this works: the core idea

The equation d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0 is the simple harmonic oscillator differential equation. Its general solution is exactly y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x. So the problem is essentially asking you to verify that the given function satisfies the equation it was designed to solve.

The key insight: differentiating sin⁡x\sin x twice gives −sin⁡x-\sin x, and differentiating cos⁡x\cos x twice gives −cos⁡x-\cos x. So the second derivative just flips the sign of each term, producing −y-y.

Step-by-step verification

1. Write down the given function

We have y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x, where AA and BB are constants.

2. Find the first derivative

Differentiate term by term:

  • Derivative of sin⁡x\sin x is cos⁡x\cos x
  • Derivative of cos⁡x\cos x is −sin⁡x-\sin x

So:

dydx=Acos⁡x−Bsin⁡x\frac{dy}{dx} = A\cos x - B\sin x

3. Find the second derivative

Differentiate dydx\frac{dy}{dx}:

  • Derivative of cos⁡x\cos x is −sin⁡x-\sin x
  • Derivative of −sin⁡x-\sin x is −cos⁡x-\cos x

So:

d2ydx2=−Asin⁡x−Bcos⁡x\frac{d^2y}{dx^2} = -A\sin x - B\cos x

4. Observe the pattern

Notice that −Asin⁡x−Bcos⁡x-A\sin x - B\cos x is exactly −(Asin⁡x+Bcos⁡x)-(A\sin x + B\cos x), which is −y-y.

Therefore:

d2ydx2=−y\frac{d^2y}{dx^2} = -y

5. Rearrange to get the required form

Adding yy to both sides:

d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0

Watch out

A common mistake is to forget the sign when differentiating cos⁡x\cos x — its derivative is −sin⁡x-\sin x, not sin⁡x\sin x. Also, when differentiating −sin⁡x-\sin x, remember the derivative is −cos⁡x-\cos x, not cos⁡x\cos x. Each sign error compounds, so check carefully.

Tip

You can verify this result instantly by remembering the pattern: for any linear combination of sin⁡x\sin x and cos⁡x\cos x, the second derivative always returns the negative of the original function. This is why d2dx2\frac{d^2}{dx^2} acts like multiplying by −1-1 on the space spanned by sin⁡x\sin x and cos⁡x\cos x.

✓Final answer

We have shown that d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0 for y=Asin⁡x+Bcos⁡xy = A\sin x + B\cos x.

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