The family of curves y = e^x(Acos x + Bsin x) | vedclass

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Question

(B) Given the 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)

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$

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(B) Given the 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)$$\frac{dy}{dx} = e^x((A+B)\cos x + (B-A)\sin x) = y + e^x(-A\sin x + B\cos x)$Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x(-A\sin x + B\cos x) + e^x(-A\cos x - B\sin x)$$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - y$$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$.

The family of curves $y = e^x(A\cos x + B\sin x)$ represents the differential equation:

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