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Q2.    Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

            (ii)    y = e^x(a\cos x + b \sin x )\qquad : \ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

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Given,

y = e^x(a\cos x + b \sin x )

Now, differentiating both sides w.r.t. x,

\frac{dy}{dx} = e^x(-a\sin x + b \cos x ) + e^x(a\cos x + b \sin x ) =e^x(-a\sin x + b \cos x ) +y

Again, differentiating both sides w.r.t. x,

\\ \frac{d^2y}{dx^2} = e^x(-a\cos x - b \sin x ) + e^x(-a\sin x + b \cos x ) + \frac{dy}{dx} \\ = -y + (\frac{dy}{dx} -y) + \frac{dy}{dx} \\ \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0

Therefore, the given function is the solution of the corresponding differential equation.

Posted by

HARSH KANKARIA

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