Question

solve the ODE: x''+5x'+4x=8

Answer 1

different process for this one

Answer 2

first lets solve x''+5x'+4x=0
r^2+5r+4=0
(r+1)(r+4)=0
r=-1, r=-4
so we have
the homogeneous solution is
\[y=c_1e^{-1t}+c_2e^{-4t}\]

Answer 3

oops that y is suppose to be x

Answer 4

and then if I have IB conditions I can just plug in and solve for C_1 and C_2?

Answer 5

and if one IB is x'(0)=0, I will also need to take the derivative of homogeneous solution?

Answer 6

we can check this
\[x'=-c_1e^{-1t}-4c_2e^{-4t}=> x''=c_1e^{-1t}+16c_2e^{-4t}\]
\[(c_1e^{-1t}+16c_2e^{-4t})+5(-c_1e^{-1t}-4c_2e^{-4t})+4(c_1e^{-1t}+c_2e^{-4t})\]
\[=(1-5+4)e^{-1t}+(16-20+4)e^{-4t}=0e^{-1t}+0e^{-4t}=0\]
so anyways
this is only part of our solution
\[x=c_1e^{-1t}+c_2e^{-4t}+t_p\]

so we need to find the non-homogeneous solution
we need to look back as
x''+5x'+4x=8

Answer 7

joe you wanna finish it?
lol

Answer 8

i cant remember the method to finding a particular solution lol >.< other than eye-balling it and saying x = 2 is a particular solution.

Answer 9

we need to find that t_p part
i think it is called the particular solution

Answer 10

lol

Answer 11

yes x=2 works

Answer 12

lolol, theres a set way of finding it, i just cant remember, ima look it up.

Answer 13

...unless phi knows it and is going to post it :)

Answer 14

x=A+Bt+Ct^2
or something like that

Answer 15

and then we find x'
and x''
and find what A B and C are

Answer 16

to make the equation true

Answer 17

lets see what phi says :)

Answer 18

You can find the particular solution by assuming a linear combination of the RHS and all its derivatives. This only works for things with a finite number of derivatives. But a constant works. Assume x= A.
x' =0 and x''=0. Plug in and solve for A
the answer is 2

Answer 19

\[x=c_1e^{-1t}+c_2e^{-4t}+2\]

Answer 20

so you said you had boundary conditions?

Answer 21

no!
stupid thing was going so slow so i clicked too many times and it deleted what i wrote :(

Answer 22

>.< lol

Answer 23

x'(0)=0, x(0)=2

Answer 24

to use x'(0)=0
find x'
\[x'=-c_1e^{-1t}-4c_2e^{-4t}\]
\[0=-c_1-4c_2\]

and to use x(0)=2
we have
\[2=c_1+c_2+2\]

Answer 25

could someone help expand on how to get the particular solution? you just set the left to be a constant A and solve x' and x''?

Answer 26

i think you have to sort of guess what the form of particular solution will be
like we could try x=A+Bt+Ct^2 and see what happens
x'=0+B+2Ct
x''=2C
so plugin' these in we get

2C+5(B+2Ct)+4(A+Bt+Ct^2)=8
we need to get our t^2 together
our t's together
and our constants

4t^2*C+t*(10C+4B)+2C+5B+4A=8

but C=0

so B=0

then 4A=8=> A=2

Answer 27

Since it is equal to a constant you can assume the particular part is a constant (or at least contains one as myininaya showed) As long as you don't guess sin(x) or cos(x) or anything. The method is called undetermined coefficients. This is a great website:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 28

how do we know C =0?

Answer 29

there are no t^2 on the other side of the equation

Answer 30

and B=0 because there are no t's either?

Answer 31

right
we have (10C+4B)t=0t
10C+4B=0
but we said first C=0
so we have 4B=0
so B=0

Answer 32

and is this A =2 our t_p?

Answer 33

There is a theorem in linear algebra that says:
\[at^n+bt^{n-1}...=\alpha t^n+\beta t^{n-1}... \implies a=\alpha; b=\beta\]
Though I gave a very "specific" by assuming it was a polynomial, it can be anything really.

Answer 34

yes dof

Answer 35

k, I think I got it; I appreciate all the time and help!

Answer 36

np
the only thing left is two solve your two equations above for c1 and c2

Answer 37

but you can do that using algebra