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find value of $K_1$ and $K_2$ for which $y= \large\frac{x}{K_1} $$+K_2$ is a solution of differential equation $y''-2y' -3y=x$

$(a)\;(-3, -\frac{2}{9}) \\ (b)\;(\frac{-2}{9},3) \\ (c)\;(-2,3) \\ (d)\;no\; real\; K_1\;and\;K_2 $
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$y= \large\frac{x}{K_1}$$+K_2$
$y'= \large\frac{1}{K_1}$
$y''=0$
Put in equation
$0- \large\frac{2}{K_1}$$-3\bigg( \large\frac{x}{K_1} $$+K_2\bigg) =x$
$\large\frac{-2}{K_1}- \frac{-3x}{K_1}$$-3K_2=x$
So, $K_1=-3$
$K_2=\large\frac{-2}{9}$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p
 

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