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2 years ago

Help to solve y'''-y''-4y'+4y=5-e^x+e^2x

Mathematics

2 answers:

Licemer1 [7]2 years ago

5 0

Answer:

Step-by-step explanation:

First of all write the auxialary equation as

$m^3-m^2-4m+4 =0\\(m-1)(m-2)(m+2)=0$

m=1,2,-2

Hence general solution is

$y=Ae^x+Be^{2x} +Ce^{-2x}$

Particular solution of 5 is

$\frac{5}{4}$

Particular solution of e^x is

$x(\frac{1}{1-4} e^x =\frac{-xe^x}{3}$

Particular solution of e^2x is

$x\frac{e^{2x} }{(2+2)(2-1)} =\frac{xe^{2x} }{4}$

Together full solution is

$y=Ae^x+Be^{2x} +Ce^{-2x}+\frac{5}{4} +\frac{xe^x}{3} +\frac{xe^{2x} }{4}$

Alecsey [184]2 years ago

4 0

y'''-y''-4y'+4y=5-ex+e2x using undetermined coefficients. I just wasted 3 hours ... hairs on my head out! Thanks in advance for any help you may provide.

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