Question

solve the given differential equation by undetermined coefficients
y'''+y"-4y'+4y=5-e^x+e^2x

Answer 1

im sorry im not sure how to do this but couldnt you combine like terms?

Answer 2

You first to find the homogeneous solution.

Answer 3

You can do that pretty easy if you know how to solve:
\[r^3+r^2-4r+4=0\]

Answer 4

That can be done by factoring by grouping.

Answer 5

Let me know if and when you have the homogeneous solution, then you can proceed to find the particular solution by the method in your post.

Answer 6

Ok for my yc I got yc =c1e^-2x+c2e^2x+c3e^x

Answer 7

Is that what you got

Answer 8

looks awesome! :)

Answer 9

now notice on the left hand side of your differential equation
you have a constant+constant*e^x+constant*e^(2x)

Answer 10

Ok

Answer 11

So we are going to take your homogeneous solution and add it to something that looks like that
so what i'm trying to say we are going to try the following as a solution:
\[y=c_1e^{-2x}+c_2e^{2x}+c_3e^{x}+c_4 \]

Answer 12

So would my yp be yp=a+bxe^x+cxe^2x

Answer 13

Ok

Answer 14

So we have to find y'
y''
y'''
and plug all of those in to find our constants

Answer 15

Ok

Answer 16

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
I really believe this site will help you to get the feel for what solutions to try.
But sometimes it is trial and error.

Answer 17

Ok but are you still going to help me

Answer 18

Yes. I was just giving the site to you because I thought it would be helpful to you to determine what things to try in the future.

Answer 19

Just let me know what else you need help on.

Answer 20

Ok so the what did you got for the final answer

Answer 21

have you plug in y^(3), y^(2),y^(1), and y yet
and then grouped your liked terms together and compared to the other sides to find out what the constant coefficients should be

Answer 22

Yes and my remainder after was -4be^x-4e^2x+4a

Answer 23

But idk what to do from there

Answer 24

\[y=c_1e^{-2x}+c_2e^{2x}+c_3e^{x}+c_4 \\ y'=-2c_1e^{-2x}+2c_2e^{2x}+c_3e^{x}+0\]
then so on with the other derivatives
c_4 should be easy to find

Answer 25

like you don't have any other constant terms in any of the other derivatives
so we know 4c_4=5

Answer 26

show me the second and third derivative
and I will show you all you need to do is plug in and collect likes and compare sides to get the constants

Answer 27

It's going to take a whil to typ can you just show me step by step of what you did

Answer 28

2 and 3rd derivative only have three terms

Answer 29

I will find the 2nd one for you
but i want you to show me you have put some work into problem

Answer 30

\[y^{(2)}=4c_1e^{-2x}+4c_2e^{2x}+c_3e^{x}\]

Answer 31

I have I'm just really lost bc of the remainder I got after lugging them all in

Answer 32

So what did you get the third derivative?

Answer 33

y"'=-8c1e^-2x+8c2e^2x+c3e^x

Answer 34

Great. I basically just wanted to make sure you weren't having problems differentiating
and clearly that isn't your trouble here.
So next part lets plug in.

Answer 35

Ok

Answer 36

\[y^{(3)}+y^{(2)}-4y^{(1)}+4y \\ =(-8c_1e^{-2x}+8c_2e^{2x}+c_3e^{x}) \\ +(4c_1e^{-2x}+4c_2e^{2x}+c_3e^{x}) \\ -4(-2c_1e^{-2x}+2c_2e^{2x}+c_3e^{x}) \\ +4(c_1e^{-2x}+c_2e^{2x}+c_3e^{x}+c_4) \]
And this side is suppose to equal the other side
\[1e^{2x}-1e^{x}+5\]

Answer 37

we only have one constant term on both sides right?

Answer 38

Yes

Answer 39

So what is next

Answer 40

so we already know that 4c_4=5 which I already said
now let's look at the e^x terms

Answer 41

group all the e^x terms together on left hand side of the equation

Answer 42

So c = 1/5

Answer 43

like this:
\[c_3e^{x}+c_3e^{x}-4c_3e^{x}+4c_3e^{x}=-1e^{x} \\ \text{ \implies } c_3+c_3-4c_3+4c_3=-1 \]

Answer 44

well c_4=5/4

Answer 45

you divide both sides by 4 to solve for c_4

Answer 46

Ok my by hahahand what is the a and b

Answer 47

what is a and b?

Answer 48

have you found c_3 yet?

Answer 49

Do you see how I got the equation involving the constant c_3?

Answer 50

Yes but I'm to nurse about it I got would it be. 1

Answer 51

Ok we have \[c_3e^{x}+c_3e^{x}-4c_3e^{x}+4c_3e^{x}=-1e^{x} \\ \text{ \implies } c_3+c_3-4c_3+4c_3=-1 \]
I know 1+1-4+4=2+0=2
so I know we have \[2c_3=-1 \]

Answer 52

I know to solve for c_3 I must divide both sides by 2

Answer 53

So it's 1/2

Answer 54

yes
all we are doing is just solving a bunch on linear equations (like the ones from algebra)

Answer 55

You try to find the constant we named c_2

Answer 56

Ok so c2 would be 1/4

Answer 57

\[y^{(3)}+y^{(2)}-4y^{(1)}+4y \\ =(-8c_1e^{-2x}+8c_2e^{2x}+c_3e^{x}) \\ +(4c_1e^{-2x}+4c_2e^{2x}+c_3e^{x}) \\ -4(-2c_1e^{-2x}+2c_2e^{2x}+c_3e^{x}) \\ +4(c_1e^{-2x}+c_2e^{2x}+c_3e^{x}+c_4) \\ =1e^{2x}-1e^{x}+5 \]
rememeber we need to compare both sides of this above:

gather all the e^{2x}'s on the left hand side just as I did before.
so first line we have 8c_2e^(2x)
second line we have +4c_2e^{2x)
third line we have -4(2c_2e^{2x))
fourth line we have +4(c_2e^(2x))
and on the other side we have 1e^(2x)
so we have know we must have the equation \[8c_2+4c_2-8c_2+4c_2=1 \]

Answer 58

Is this what you tried to solve?

Answer 59

So. 1/8

Answer 60

yes :)

Answer 61

now last constant c_1

Answer 62

Ci is 1/8 as wel

Answer 63

be careful with that

Answer 64

do we have any e^(-2x) 's on the other side?

Answer 65

\[(-8c_1+4c_1-8c_1+4c_1)e^{-2x}=0e^{-2x} \\ 8c_1=0 \]
We did not did we ?

Answer 66

remember we have 5-e^x+e^(2x)

Answer 67

that didn't consist of any e^{-2x) terms
so here we find out we didn't even need to include the c_1e^(-2x) in our solution

Answer 68

we chose this as our solution and we found the constants\[y=c_1e^{-2x}+c_2e^{2x}+c_3e^{x}+c_4 \]
so last step is to replace those constants with what we found

Answer 69

Son 0

Answer 70

yes c1 is 0

Answer 71

\[y=c_2e^{2x}+c_3e^{x}+c_4 \]
but you did find constant values for these other c's that weren't 0

Answer 72

moos the final set Iwould e wh.

Answer 73

So what wod the yin Laing look like and please Jung tell me c Inge do work on home therhtuff