The differential equation whose solution is y | vedclass

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Question

(B) Given the solution is $y = c_1 \cos ax + c_2 \sin ax$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = -c_1 a \sin ax + c_2 a \cos ax

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

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(B) Given the solution is $y = c_1 \cos ax + c_2 \sin ax$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = -c_1 a \sin ax + c_2 a \cos ax$. Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = -c_1 a^2 \cos ax - c_2 a^2 \sin ax$. Factoring out $-a^2$: $\frac{d^2y}{dx^2} = -a^2(c_1 \cos ax + c_2 \sin ax)$. Since $y = c_1 \cos ax + c_2 \sin ax$,we substitute $y$ into the equation: $\frac{d^2y}{dx^2} = -a^2 y$. Rearranging the terms,we get: $\frac{d^2y}{dx^2} + a^2 y = 0$.

The differential equation whose solution is $y = c_1 \cos ax + c_2 \sin ax$ is (where $c_1, c_2$ are arbitrary constants):

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