The differential equation of the family of cu | vedclass

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(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$Differentiating with respect to $x$:$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B

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

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(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$Differentiating with respect to $x$:$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$$\frac{dy}{dx} = y + e^{x}(-A \sin x + B \cos x)$ ... $(2)$Differentiating again with respect to $x$:$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)$Substitute $e^{x}(-A \sin x + B \cos x) = \frac{dy}{dx} - y$ from $(2)$:$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^{x}(A \cos x + B \sin x)$Since $e^{x}(A \cos x + B \sin x) = y$:$\frac{d^{2}y}{dx^{2}} = 2\frac{dy}{dx} - y - y$$\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0$

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$,where $A$ and $B$ are arbitrary constants,is

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