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

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

  • $\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,
$\Rightarrow (a+b)[e^x(-\sin x)+ e^x.\cos x] + (b - a)[e^x\cos x + e^x\sin x]$
$\large\frac{d^2y}{dx^2 }$$= - ae^x\sin x + ae^x \cos x - be^x \sin x +be^x\cos x + be^x\cos x +be^x\sin x - ae^x\cos x -ae^x\sin x$
On simplifying we get,
$2e^x[b\cos x - a\sin x]$
Step 2:
substituting for $\large\frac{d^2y}{dx^2} ,\frac{dy}{dx}$ and $y$ in the solution we get,
$2e^x[b\cos x - a\sin x] - {2[(a+b)e^x\cos x + (b - a) e^x\sin x ]}+2e^x[a\cos x + b\sin x]$
$2be^x\cos x -2ae^x\sin x -2ae^x\cos x -2be^x\cos x - 2be^x\sin x + 2ae^x\sin x + 2ae^x\cos x + 2be^x\sin x = 0$
Hence the given function is a solution to the given differential equation.
answered Jul 30, 2013 by sreemathi.v

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