How to find the general solution to y"'+2y"-4y'-8y=0?

asked Jul 7, 2018 14.6k views

3 votes

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4 votes

has characteristic equation

r^3+2r^2-4r-8=r^2(r+2)-4(r+2)=(r^2-4)(r+2)=(r-2)(r+2)^2=0

which has roots at
$r=\pm2$ . The negative root has multiplicity 2. So the general solution is

y=C_1e^(2x)+C_2e^(-2x)+C_3xe^(-2x)

Milander answered Jul 13, 2018

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