If y = sin ax + cos bx, then y'' + b^2 y = | vedclass

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(A) Given $y = \sin ax + \cos bx$. First, differentiate with respect to $x$: $y' = \frac{d}{dx}(\sin ax) + \frac{d}{dx}(\cos bx) = a \cos ax - b \sin

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$(b^2 - a^2) \sin ax$

Vedclass · Answer

(A) Given $y = \sin ax + \cos bx$. First, differentiate with respect to $x$: $y' = \frac{d}{dx}(\sin ax) + \frac{d}{dx}(\cos bx) = a \cos ax - b \sin bx$. Next, differentiate again to find the second derivative: $y'' = \frac{d}{dx}(a \cos ax - b \sin bx) = -a^2 \sin ax - b^2 \cos bx$. Now, substitute $y''$ and $y$ into the expression $y'' + b^2 y$: $y'' + b^2 y = (-a^2 \sin ax - b^2 \cos bx) + b^2(\sin ax + \cos bx)$. $y'' + b^2 y = -a^2 \sin ax - b^2 \cos bx + b^2 \sin ax + b^2 \cos bx$. $y'' + b^2 y = (b^2 - a^2) \sin ax$.

If $y = \sin ax + \cos bx$, then $y'' + b^2 y =$

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