The differential equation whose solution is y | vedclass

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Question

(A) Given the equation $y = A\sin x + B\cos x$. Differentiating with respect to $x$ once,we get: $\frac{dy}{dx} = A\cos x - B\sin x$. Differentiating

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

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(A) Given the equation $y = A\sin x + B\cos x$. Differentiating with respect to $x$ once,we get: $\frac{dy}{dx} = A\cos x - B\sin x$. Differentiating with respect to $x$ again,we get: $\frac{d^2y}{dx^2} = -A\sin x - B\cos x$. We can rewrite this as: $\frac{d^2y}{dx^2} = -(A\sin x + B\cos x)$. Since $y = A\sin x + B\cos x$,we substitute $y$ into the equation: $\frac{d^2y}{dx^2} = -y$. Rearranging the terms,we get the required differential equation: $\frac{d^2y}{dx^2} + y = 0$.

The differential equation whose solution is $y = A\sin x + B\cos x$ is

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