The differential equation of y=e^x(a cos x+b | vedclass

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(D) Given equation: $y = e^x(a \cos x + b \sin x)$ Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx} = e^x(a \cos x + b \sin

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

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(D) Given equation: $y = e^x(a \cos x + b \sin x)$ Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$ Since $y = e^x(a \cos x + b \sin x)$,we can write: $\frac{dy}{dx} = y + e^x(b \cos x - a \sin x)$ Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(b \cos x - a \sin x) + e^x(-b \sin x - a \cos x)]$ Substitute $e^x(b \cos x - a \sin x) = \frac{dy}{dx} - y$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^x(a \cos x + b \sin x)$ Since $e^x(a \cos x + b \sin x) = y$,we have: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{dy}{dx} - y - y$ $\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$ Rearranging the terms: $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

The differential equation of $y=e^x(a \cos x+b \sin x)$ is

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