Question

Solve the initial value problem
y"+4y'+3y = e^(-x)
with y(0) = 2 and y'(0) = 1

Answer 1

have you solved for the homogeneous solution yet?

Answer 2

The first part is y=Ae^-x + Be^-3x

Answer 3

ok good lol

Answer 4

what do i use for the partial integral?

Answer 5

so now we have to worry about the particular solution

Answer 6

so what i what do is do undetermined coefficients

Answer 7

let me try

Answer 8

i tried y=\[\alpha\]e^-x

Answer 9

but kept getting zero as my answer!

Answer 10

bookmarking for future reference.

Answer 11

If the right side of your equation (in this case e^(-x) ) is found in the homogeneous solution, then when you use the method of undetermined coefficients I believe you need to multiply the new term of your solution by a factor of x. That way your solutions will be linearly independent. So instead of trying Ce^(-x), try Cxe^(-x). Correct me if I'm wrong, it's been a while since I've done these.

Answer 12

\[y=A+Be^{x}+Ce^{-x}+Dxe^{x}+Exe^{-x}\]
\[y'=Be^x-Ce^{-x}+De^x+Dxe^{x}+Ee^{-x}-Exe^{-x}\]
\[y''=Be^{x}+Ce^{-x}+De^{x}+De^x+Dxe^x-Ee^{-x}+Exe^{-x}-Ee^{-x}\]
------------------------------------------------------------------
plug in:
\[Be^{x}+Ce^{-x}+De^{x}+De^x+Dxe^x-Ee^{-x}+Exe^{-x}-Ee^{-x}\]
\[+4(Be^x-Ce^{-x}+De^x+Dxe^{x}+Ee^{-x}-Exe^{-x})\]
\[+3(A+Be^x+Ce^{-x}+Dxe^{x}+Exe^{-x}\]
=
(note: lets group our like terms together)
\[3A+e^{x}(B+D+D+4B+4D+3B)+e^{-x}(C-E-E-4C+4E)\]
\[+xe^{x}(D+4D+3D)+xe^{-x}(E-4E+3E)\]
=
\[e^{-x}\]

so we have
A=0

8B+6D=0

-3C-2E=-1

8D=0

0E=0

Answer 13

this is what i have so far
i'm pretty sure i made a mistake
it still putting everything on one line lol

Answer 14

one second let me write that last line better

Answer 15

\[A=0\]
\[8B+6D=0\]
\[-3C-2E=-1\]
\[8D=0\]
\[0E=0\]

Answer 16

why do we need so many undetermined parameters?

Answer 17

because i don't know what solution form we have exactly

Answer 18

but anyways

if D=0, then B=0, A=0

so we just need to look at C and E

Answer 19

it looks like E is any constant

Answer 20

or can be any constant

Answer 21

unless i made a mistake above

Answer 22

so if C can be any constant then
\[-3C=-1+2E => C=\frac{1-2E}{3}\]

Answer 23

looks good to me im gonna write it all out

Answer 24

thanks alot i'll try and finalise the answer and post back

Answer 25

yea i couldn't see any mistakes
its weird though that we got E can by any constant

Answer 26

cow what do you think?
do you see any error?

Answer 27

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 28

ok so this site is saying that the particular solution should be in this form:
\[y=Ae^{-x}\]

Answer 29

\[y'=-Ae^{-x}=>y''=Ae^{-x}\]
\[Ae^{-x}+4(-Ae^{-x})+3(Ae^{-x})=e^{-x}\]
\[(A-4A+3A)e^{-x}=e^{-x}=> 0Ae^{-x}=e^{-x}\]
this is not working

Answer 30

no i think i got it. PS should be y=Axe^-x

Answer 31

lol
i tried so hard :(

Answer 32

therefore y'=-Axe^-x + Ae^-x

Answer 33

myininaya, look at example 9 on the site you posted. That's similar to the problem we're dealing with here.

Answer 34

and y''=Axe^-x -2Ae^-x

Answer 35

solving gives A=1/2

Answer 36

ok ok i see

Answer 37

so general solution: y=Ae^-x +Be^-3x +1/2xe^-x

Answer 38

ok cool

Answer 39

Actually I think the general solution is y=Ae^-x +Be^-3x +Cxe^-x

Answer 40

c being 1/2

Answer 41

I'm lost, how'd you get c = 1/2 ?

Answer 42

lets see
\[y=Axe^{-x}=>y'=Ae^{-x}-Axe^{-x}=>y''=-Ae^{-x}-Ae^{-x}+Axe^{-x}\]

Answer 43

\[(-Ae^{-x}-Ae^{-x}+Axe^{-x})+4(Ae^{-x}-Axe^{-x})+3(Axe^{-x})=e^{-x}\]

Answer 44

so we have
\[xe^{-x}(A-4A+3A)+e^{-x}(-A-A+4A)=e^{-x}\]

Answer 45

you put in y, y' and y'' into the auxillary eq. giving you Cxe^-x -2Ce^-x -aCxe^-x +4Ce^-x +3Cxe^-x = e^-x

Answer 46

\[2A=1\]

Answer 47

\[A=\frac{1}{2}\]

Answer 48

yeah thats right

Answer 49

is everyone good?

Answer 50

i think we are good now

Answer 51

with y(0) = 2 and y'(0) = 1 the equation i think is 11/4e^-x -3/4e^-3x +1/2xe^-x

Answer 52

ok so after using you intial conditions..
we have y(0)=1 and y'(0)=1..
\[y=Ae^{-x}+Be^{-3x}+\frac{1}{2}xe^{-x}\]
\[y(0)=2=A+B, y'(0)=1=-A-3B+\frac{1}{2}\]
is that right?

Answer 53

no not +1/2 cos 1/2times0 =0

Answer 54

but the rest is right

Answer 55

\[A=2-B=> 1=-(2-B)-3B+\frac{1}{2}=>1=-2-2B+\frac{1}{2}=>3-\frac{1}{2}=-2B\]
well
\[y'=-Ae^{-x}-3e^{-3x}+\frac{1}{2}e^{-x}-\frac{1}{2}xe^{-x}\]

Answer 56

\[y'(0)=1=-A-3B+\frac{1}{2}\]

Answer 57

yes i know i missed my B

Answer 58

I think myininaya is right

Answer 59

\[\frac{5}{2}=-2B =>B=\frac{-5}{4}\]

Answer 60

how did u get 1/2e^-x - 1/2xe^-x

Answer 61

\[A=2-(\frac{-5}{4})=2+\frac{5}{4}=\frac{9}{4}\]

Answer 62

ok we have
\[y=Ae^{-x}+Be^{-3x}+\frac{1}{2}xe^{-x}\]
right?

Answer 63

u know the y' i think its only supposed to be y' of the y you have found now. not anything to do with the particular soln

Answer 64

so you wouldnt add the +1/2e^-x

Answer 65

oh you would. ok ok i get it! lol sorry

Answer 66

i was trying to take shortcuts. thanks for everyones help

Answer 67

np
i learned something today
if you are getting 0 multiply by x lol

Answer 68

ok for some reason my openstudy is freaking out and going really slow

Answer 69

is it possible you could type the full answer out so i can just check for sure please

Answer 70

\[y=Ae^{-x}+Be^{-3x}+\frac{1}{2}xe^{-x} \]

Answer 71

A=9/4, B=-5/4

Answer 72

\[y=\frac{9}{4}e^{-x}+\frac{-5}{4}e^{-3x}+\frac{1}{2}xe^{-x}\]

Answer 73

thanks so much!