Question

If \( y = A \sin x + B \cos x \), then find the differential equation of it.

Hint:

The differential equation for a linear combination of sine and cosine functions is always of the form \( \frac{d^2y}{dx^2} + y = 0 \).

Answer 1

We are given that \[ y = A \sin x + B \cos x. \] Now, differentiate both sides of the equation with respect to \( x \): \[ \frac{dy}{dx} = A \cos x - B \sin x. \] Next, differentiate the first derivative with respect to \( x \): \[ \frac{d^2y}{dx^2} = -A \sin x - B \cos x. \] Now, observe that the second derivative is equal to the negative of the original function \( y \): \[ \frac{d^2y}{dx^2} = -y. \] Thus, the required differential equation is: \[ \frac{d^2y}{dx^2} + y = 0. \] Conclusion: The differential equation of \( y = A \sin x + B \cos x \) is \[ \frac{d^2y}{dx^2} + y = 0. \]