Question

If \(y = \sin ax + \cos bx\) then \(y'' + b^2y =\)

• A. \((b^2 - a^2)\sin ax\)
• B. \((b^2 - a^2)\cos bx\)
• C. \((a^2 - b^2)\tan ax\)
• D. \((b^2 - a^2)\cot bx\)

Hint:

Remember the trigonometric identities for derivatives, especially when working with linear combinations of sine and cosine functions.

Answer 1

The Correct Option is A

Given \(y = \sin ax + \cos bx\), the second derivative \(y''\) is calculated as follows: \[ y' = a \cos ax - b \sin bx \] \[ y'' = -a^2 \sin ax - b^2 \cos bx \] Adding \(b^2y\) to \(y''\): \[ y'' + b^2y = (-a^2 \sin ax - b^2 \cos bx) + b^2(\sin ax + \cos bx) \] \[ = -a^2 \sin ax \] This gives the expression \((b^2 - a^2)\sin ax\), confirming the correct answer.