The differential equation of the family of cu | vedclass

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(B) Given equation: $y = e^{x}(A \cos x + B \sin x)$ Differentiating with respect to $x$ using the product rule: $y' = e^{x}(A \cos x + B \sin x) + e^

Vedclass · Accepted Answer

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

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(B) Given equation: $y = e^{x}(A \cos x + B \sin x)$ Differentiating with respect to $x$ using the product rule: $y' = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$ $y' = y + e^{x}(-A \sin x + B \cos x)$ Differentiating again with respect to $x$: $y'' = y' + [e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)]$ From the first derivative, we know $e^{x}(-A \sin x + B \cos x) = y' - y$. Substituting this into the second derivative: $y'' = y' + (y' - y) - e^{x}(A \cos x + B \sin x)$ Since $y = e^{x}(A \cos x + B \sin x)$, we have: $y'' = 2y' - y - y$ $y'' - 2y' + 2y = 0$ Thus, the differential equation is $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$.

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

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