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Worked Examples · Example 3

Q.Verify that the function y=acos⁡x+bsin⁡xy = a\cos x + b\sin x, where a,b∈Ra, b \in \mathbf{R} is a solution of the differential equation d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0.

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The function y=acos⁡x+bsin⁡xy = a\cos x + b\sin x satisfies d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0 because its second derivative is −y-y, making the sum identically zero for any real a,ba, b.

Why this approach works

When we're asked to verify that a given function is a solution of a differential equation, we don't need to solve anything — we just need to check that plugging the function into the equation makes it true. The differential equation here is second-order linear and homogeneous: d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0. It says, in words, "the second derivative of yy plus yy itself equals zero for all xx."

The given function is a combination of sine and cosine. The key insight: both sin⁡x\sin x and cos⁡x\cos x have the property that differentiating them twice brings you back to the negative of the original function. So any linear combination of them will also have that property. That's exactly what the equation demands.


Step-by-step verification

1. Write down the function clearly

We have:

y=acos⁡x+bsin⁡xy = a\cos x + b\sin x

where aa and bb are any real constants.

2. Find the first derivative

Differentiate term by term:

dydx=−asin⁡x+bcos⁡x\frac{dy}{dx} = -a\sin x + b\cos x

Remember: derivative of cos⁡x\cos x is −sin⁡x-\sin x, and derivative of sin⁡x\sin x is cos⁡x\cos x.

3. Find the second derivative

Differentiate dydx\frac{dy}{dx}:

d2ydx2=−acos⁡x−bsin⁡x\frac{d^2y}{dx^2} = -a\cos x - b\sin x

Here, derivative of −sin⁡x-\sin x is −cos⁡x-\cos x, and derivative of cos⁡x\cos x is −sin⁡x-\sin x.

4. Observe the pattern

Look at the second derivative:

d2ydx2=−(acos⁡x+bsin⁡x)\frac{d^2y}{dx^2} = -(a\cos x + b\sin x)

But acos⁡x+bsin⁡xa\cos x + b\sin x is exactly yy! So:

d2ydx2=−y\frac{d^2y}{dx^2} = -y …

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