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Verified Answer :

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

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

Cricetus

By solving the equation "\(y'''-y''-4y'+4y=5-e^x+e^2x\)" we get "\(y = Ae^x+Be^{2e}+Ce^{-2x}+\frac{5}{4}+\frac{xe^x}{3} +\frac{xe^{2x}}{4}\)".

As we know the auxiliary equations,

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

here,

m = 1, 2, -2

Thus,

The general equation will be:

→ \(y = Ae^x+Be^{2x}+Ce^{-2x}\)

Particularly,

  • \(5 = \frac{5}{4}\)
  • \(e^x= x(\frac{1}{1-4}e^x ) = -\frac{xe^x}{3}\)
  • \(e^{2x} = x(\frac{e^{2x}}{(2+2)(2-1)} )= \frac{xe^{2x}}{4}\)

hence,

The complete equation will be:

→ \(y = Ae^x+Be^{2e}+Ce^{-2x}+\frac{5}{4}+\frac{xe^x}{3} +\frac{xe^{2x}}{4}\)

Thus the above answer is right.

Learn more:

https://brainly.com/question/22449567

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