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ghdoru11

  • one year ago

compute y(2) if y(0)=y'(0)=0 and y''-3y'+2y=xe^x

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  1. ghdoru11
    • one year ago
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    @SithsAndGiggles @beginnersmind @IrishBoy123 @Loser66 some of you can help me with this problem? i'll be thankful

  2. IrishBoy123
    • one year ago
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    i'm in i'd give it a good hard slap with a Laplace Transform, but there are probably other ways too.

  3. ghdoru11
    • one year ago
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    i tried using the homogenous solution and the particular solution but i can't figure it out

  4. IrishBoy123
    • one year ago
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    so first you solved \[ y''-3y'+2y=0\] and you got an answer?

  5. IrishBoy123
    • one year ago
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    then you turned your mind to \[y''-3y'+2y=x \ e^x\]

  6. IrishBoy123
    • one year ago
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    for 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

  7. ghdoru11
    • one year ago
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    i don't know if it is right or not

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  8. IrishBoy123
    • one year ago
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    has 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

  9. IrishBoy123
    • one year ago
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    another way, a bit quicker

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  10. Jhannybean
    • one year ago
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    Your writing is a little messy!

  11. IrishBoy123
    • one year ago
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    it's deliberate! it's all a load of rubbish really, i just made it up and posted it :-))

  12. ghdoru11
    • one year ago
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    @IrishBoy123 all you done there is the solve or just a part of it?

  13. IrishBoy123
    • one year ago
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    attachment 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.

  14. ghdoru11
    • one year ago
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    thank you for your time, now it's clear to me. hope i will pass the exam

  15. IrishBoy123
    • one year ago
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    if 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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