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

Answers

Answer 1

Answer:

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

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,

$5 = (5)/(4)$
$e^x= x((1)/(1-4)e^x ) = -(xe^x)/(3)$
$e^(2x) = x((e^(2x))/((2+2)(2-1)) )= (xe^(2x))/(4)$

hence,

The complete equation will be:

→ $y = Ae^x+Be^(2e)+Ce^(-2x)+(5)/(4)+(xe^x)/(3) +(xe^(2x))/(4)$

Thus the above answer is right.

Learn more:

brainly.com/question/22449567

Answer 2

Answer:

Answer:

Step-by-step explanation:

First of all write the auxialary equation as

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

m=1,2,-2

Hence general solution is

$y=Ae^x+Be^(2x) +Ce^(-2x)$

Particular solution of 5 is

$(5)/(4)$

Particular solution of e^x is

$x((1)/(1-4) e^x =(-xe^x)/(3)$

Particular solution of e^2x is

$x(e^(2x) )/((2+2)(2-1)) =(xe^(2x) )/(4)$

Together full solution is

$y=Ae^x+Be^(2x) +Ce^(-2x)+(5)/(4) +(xe^x)/(3) +(xe^(2x) )/(4)$

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Answers

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Math Graded Assignment Unit Test, Part 2 Measures of Center and Spread(Score for Question 2: ___ of 5 points)
2. Consider the following line plot.
2
4
6
8
(a) What is the general trend of the graph?
(b) What is the median of the data? Explain.
(c) What is the mean of the data? Explain. Round to the Nearest tenth.
(d) Would the mean or median be affected more with a data point of 20? Explain.
Answer:
P