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The P.L of $(3D^{2}+D-14)y=13e^{2x}$ is

\[\begin{array}{1 1}(1)26x\;e^{2x}&(2)13x\;e^{2x}\\(3)x\;e^{2x}&(4)x^{2}/2\;e^{2x}\end{array}\]

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$(3D^2+D-14)y =13e^{2x}$
$P.I= \large\frac{13 e^{2x}}{3D^2+7D- 6D-14}$
$\qquad= \large\frac{13 e^{2x}}{3D^2+7D-6D-14}$
$\qquad=\large\frac{13 e^{2x}}{D(3D+7) -2(3D+7)}$
$\qquad= \large\frac{13}{6+7}. \frac{e^{2x}}{D-2} $
$\qquad= xe^{2x}$
Hence 3 is the correct answer.
answered May 22, 2014 by meena.p
 

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