Question

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

Answer 1

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

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

Answer 2

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.

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