The differential equation formed by eliminati | vedclass

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

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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 is $y=e^x(a \cos x+b \sin x)$.Differentiating with respect to $x$ using the product rule:$\frac{d y}{d x} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$$\frac{d y}{d x} = y + e^x(-a \sin x + b \cos x) \quad \dots(I)$Differentiating again with respect to $x$:$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + \frac{d}{d x}[e^x(-a \sin x + b \cos x)]$$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)$From equation $(I)$,$e^x(-a \sin x + b \cos x) = \frac{d y}{d x} - y$.Substituting this into the second derivative:$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + (\frac{d y}{d x} - y) - e^x(a \cos x + b \sin x)$Since $e^x(a \cos x + b \sin x) = y$,we have:$\frac{d^2 y}{d x^2} = 2 \frac{d y}{d x} - y - y$$\frac{d^2 y}{d x^2} - 2 \frac{d y}{d x} + 2 y = 0$.

The differential equation formed by eliminating $a$ and $b$ from the equation $y=e^x(a \cos x+b \sin x)$ is

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