Mathematics OpenStudy (anonymous):

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

have you solved for the homogeneous solution yet? OpenStudy (anonymous):

The first part is y=Ae^-x + Be^-3x myininaya (myininaya):

ok good lol OpenStudy (anonymous):

what do i use for the partial integral? myininaya (myininaya):

so now we have to worry about the particular solution myininaya (myininaya):

so what i what do is do undetermined coefficients myininaya (myininaya):

let me try OpenStudy (anonymous):

i tried y=$\alpha$e^-x OpenStudy (anonymous):

but kept getting zero as my answer! OpenStudy (anonymous):

bookmarking for future reference. OpenStudy (anonymous):

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. myininaya (myininaya):

$y=A+Be^{x}+Ce^{-x}+Dxe^{x}+Exe^{-x}$ $y'=Be^x-Ce^{-x}+De^x+Dxe^{x}+Ee^{-x}-Exe^{-x}$ $y''=Be^{x}+Ce^{-x}+De^{x}+De^x+Dxe^x-Ee^{-x}+Exe^{-x}-Ee^{-x}$ ------------------------------------------------------------------ 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 myininaya (myininaya):

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

one second let me write that last line better myininaya (myininaya):

$A=0$ $8B+6D=0$ $-3C-2E=-1$ $8D=0$ $0E=0$ OpenStudy (anonymous):

why do we need so many undetermined parameters? myininaya (myininaya):

because i don't know what solution form we have exactly myininaya (myininaya):

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

it looks like E is any constant myininaya (myininaya):

or can be any constant myininaya (myininaya):

unless i made a mistake above myininaya (myininaya):

so if C can be any constant then $-3C=-1+2E => C=\frac{1-2E}{3}$ OpenStudy (anonymous):

looks good to me im gonna write it all out OpenStudy (anonymous):

thanks alot i'll try and finalise the answer and post back myininaya (myininaya):

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

cow what do you think? do you see any error? myininaya (myininaya): myininaya (myininaya):

ok so this site is saying that the particular solution should be in this form: $y=Ae^{-x}$ myininaya (myininaya):

$y'=-Ae^{-x}=>y''=Ae^{-x}$ $Ae^{-x}+4(-Ae^{-x})+3(Ae^{-x})=e^{-x}$ $(A-4A+3A)e^{-x}=e^{-x}=> 0Ae^{-x}=e^{-x}$ this is not working OpenStudy (anonymous):

no i think i got it. PS should be y=Axe^-x myininaya (myininaya):

lol i tried so hard :( OpenStudy (anonymous):

therefore y'=-Axe^-x + Ae^-x OpenStudy (anonymous):

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

and y''=Axe^-x -2Ae^-x OpenStudy (anonymous):

solving gives A=1/2 myininaya (myininaya):

ok ok i see OpenStudy (anonymous):

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

ok cool OpenStudy (anonymous):

Actually I think the general solution is y=Ae^-x +Be^-3x +Cxe^-x OpenStudy (anonymous):

c being 1/2 OpenStudy (anonymous):

I'm lost, how'd you get c = 1/2 ? myininaya (myininaya):

lets see $y=Axe^{-x}=>y'=Ae^{-x}-Axe^{-x}=>y''=-Ae^{-x}-Ae^{-x}+Axe^{-x}$ myininaya (myininaya):

$(-Ae^{-x}-Ae^{-x}+Axe^{-x})+4(Ae^{-x}-Axe^{-x})+3(Axe^{-x})=e^{-x}$ myininaya (myininaya):

so we have $xe^{-x}(A-4A+3A)+e^{-x}(-A-A+4A)=e^{-x}$ OpenStudy (anonymous):

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 myininaya (myininaya):

$2A=1$ myininaya (myininaya):

$A=\frac{1}{2}$ OpenStudy (anonymous):

yeah thats right myininaya (myininaya):

is everyone good? myininaya (myininaya):

i think we are good now OpenStudy (anonymous):

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

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? OpenStudy (anonymous):

no not +1/2 cos 1/2times0 =0 OpenStudy (anonymous):

but the rest is right myininaya (myininaya):

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

$y'(0)=1=-A-3B+\frac{1}{2}$ myininaya (myininaya):

yes i know i missed my B OpenStudy (anonymous):

I think myininaya is right myininaya (myininaya):

$\frac{5}{2}=-2B =>B=\frac{-5}{4}$ OpenStudy (anonymous):

how did u get 1/2e^-x - 1/2xe^-x myininaya (myininaya):

$A=2-(\frac{-5}{4})=2+\frac{5}{4}=\frac{9}{4}$ myininaya (myininaya):

ok we have $y=Ae^{-x}+Be^{-3x}+\frac{1}{2}xe^{-x}$ right? OpenStudy (anonymous):

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 OpenStudy (anonymous):

so you wouldnt add the +1/2e^-x OpenStudy (anonymous):

oh you would. ok ok i get it! lol sorry OpenStudy (anonymous):

i was trying to take shortcuts. thanks for everyones help myininaya (myininaya):

np i learned something today if you are getting 0 multiply by x lol myininaya (myininaya):

ok for some reason my openstudy is freaking out and going really slow OpenStudy (anonymous):

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

$y=Ae^{-x}+Be^{-3x}+\frac{1}{2}xe^{-x}$ myininaya (myininaya):

A=9/4, B=-5/4 myininaya (myininaya):

$y=\frac{9}{4}e^{-x}+\frac{-5}{4}e^{-3x}+\frac{1}{2}xe^{-x}$ OpenStudy (anonymous):

thanks so much!

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