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vaieri [72.5K]
4 years ago
13

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

Mathematics
1 answer:
S_A_V [24]4 years ago
8 0
y'''+2y''-4y'-8y=0

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}
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The line passes thru (6,4).  We can immediately write the desired equation as follows:

y - 4 = (-3/2)(x-6).  This could, of course, be written differently:

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6 0
3 years ago
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Elena-2011 [213]

Answer:

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3 years ago
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ICE Princess25 [194]

Answer:

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3 0
4 years ago
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KATRIN_1 [288]

Answer:

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