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$y=e^{(t+2)x}$ is a solution of equation $\large\frac{d^2y}{dx^2}-\frac{4dy}{dx}$$+4y=0$ then t equals

$(a)\;no \;real\;t \\ (b)\;1 \\ (c)\;0 \\ (d)\;2 $
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$y'= (t+2)e^{(t+2)x}$
$y''= (t+2)^2 e^{(t+2)x}$
$(t+2)^2 e^{(t+2)x}-4 (t+2)e^{(t+2)x}+4e^{(t+2)x}=0$
$(t+2)^2-4 (t+2)+4=0$
$t^2+4+4t-8+4=0$
$t^2=0$
$t=0$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
 

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