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

have you solved for the homogeneous solution yet?

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

ok good lol

what do i use for the partial integral?

so now we have to worry about the particular solution

so what i what do is do undetermined coefficients

let me try

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

but kept getting zero as my answer!

bookmarking for future reference.

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.

\[y=A+Be^{x}+Ce^{-x}+Dxe^{x}+Exe^{-x}\] \[y'=Be^x-Ce^{-x}+De^x+Dxe^{x}+Ee^{-x}-Exe^{-x}\] \[y''=Be^{x}+Ce^{-x}+De^{x}+De^x+Dxe^x-Ee^{-x}+Exe^{-x}-Ee^{-x}\] ------------------------------------------------------------------ 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

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

one second let me write that last line better

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

why do we need so many undetermined parameters?

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

but anyways if D=0, then B=0, A=0 so we just need to look at C and E

it looks like E is any constant

or can be any constant

unless i made a mistake above

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

looks good to me im gonna write it all out

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

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

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

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

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

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

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

lol i tried so hard :(

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

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

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

solving gives A=1/2

ok ok i see

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

ok cool

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

c being 1/2

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

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

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

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

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

\[2A=1\]

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

yeah thats right

is everyone good?

i think we are good now

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

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?

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

but the rest is right

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

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

yes i know i missed my B

I think myininaya is right

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

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

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

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

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

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

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

i was trying to take shortcuts. thanks for everyones help

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

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

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

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

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

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

thanks so much!

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