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

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

Toolbox:
• $\large\frac{d(e^x)}{dx }$$= e^x; \large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0. The product rule of differentiation states uv = uv' + vu' Step 1: Given : y = e^x.a\cos x + e^xb\sin x Differentiating on both sides with respect to x we get, \large\frac{dy}{dx} =$$ e^x.(-a\sin x) + e^x.a\cos x + e^x.b\cos x + e^x.b\sin x$
$\quad\;\;= (a+b)e^x\cos x + (b - a)e^x\sin x$
Differentiating again with respect to $x$ we get,
$\Rightarrow (a+b)[e^x(-\sin x)+ e^x.\cos x] + (b - a)[e^x\cos x + e^x\sin x]$

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