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anonymous
 one year ago
compute y(2) if y(0)=y'(0)=0 and y''3y'+2y=xe^x
anonymous
 one year ago
compute y(2) if y(0)=y'(0)=0 and y''3y'+2y=xe^x

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles @beginnersmind @IrishBoy123 @Loser66 some of you can help me with this problem? i'll be thankful

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i'm in i'd give it a good hard slap with a Laplace Transform, but there are probably other ways too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried using the homogenous solution and the particular solution but i can't figure it out

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so first you solved \[ y''3y'+2y=0\] and you got an answer?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2then you turned your mind to \[y''3y'+2y=x \ e^x\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2for that particular solution, i'd try: \[y = (Px^2 + Qx + R) e^x \] normally: \(y = (Px + Q) e^x \) would be good but you already have an \(e^x\) term in the complementary solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know if it is right or not

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2has this gone a bit skewy at the end? in terms of the IV's you should be putting those into the whole solution, not just the particular, unless i am misreading in terms of method, this is "Variation of Parameters"(??), because there is a really streamlined process on Paul's Online [together with a proof] it looks a lot like what you have done but with fewer steps so the Wronskian \[\left\begin{matrix}e^{2x} & e^x \\ 2e^{2x} & e^x\end{matrix}\right =  e^{3x}\] which gives \[y_p = e^{2x} \int \frac{e^x.xe^x }{ e^{3x} } \ dx + e^{x} \int \frac{e^{2x}.xe^x }{ e^{3x} } \ dx \] which i reckon comes out as \[y_p = xe^x  \frac{x^2e^x}{2}e^x\] so \[y = xe^x  \frac{x^2e^x}{2}e^x + A e^{2x} + Be^x\] and put IV's in there for A = 1, B = 0 i've done a transform and i get the same thing link to Paul: http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2another way, a bit quicker

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your writing is a little messy!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2it's deliberate! it's all a load of rubbish really, i just made it up and posted it :))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 all you done there is the solve or just a part of it?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2attachment is solution using transform, comments above are just a few comments on the way you are doing it. the solutions on Paul's seem very compact.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you for your time, now it's clear to me. hope i will pass the exam

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2if you're bothered, cover the solutions to the examples on Paul's Online and try them yourself good luck with everything!!!
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