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A particular integral of $(D^{2}-4D+4)y=e^{2x}$ is

\[\begin{array}{1 1}(1)\frac{x^{2}}{2}e^{2x}&(2)xe^{2x}\\(3)xe^{-2x}&(4)\frac{x}{2}e^{-2x}\end{array}\]

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1 Answer

$(D^2-4D +4)y=e^{2x}$
$P.I= \large\frac{e^{2x}}{D^2 -4D +4}$
$\qquad= \large\frac{e^{2x}}{(D-2)^2}=\frac{x^2}{2}$$e^{2x}$
Hence 1 is the correct answer.
answered May 21, 2014 by meena.p
 

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