# Find all values of m for which $y-e^{mx}$ is solution of equation $y'''-2y''-y'+2y=0$

$(a)\;1,2,3 \\ (b)\;1,-1,0 \\ (c)\;1,-1,2\\ (d)\;1,0,2$

## 1 Answer

$y'= me^{mx}$
$y''=m^2 e^{mx}$
$y'''=m^3e^{mx}$
$m^3-2 m^2-m+2=0$
$(m-1)(m^2-m-2)=0$
$(m-1)((m-2)(m+1))=0$
$m=1, -1,2$
Hence c is the correct answer
answered Feb 7, 2014 by

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