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anonymous

  • one year ago

check my work ??

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  1. anonymous
    • one year ago
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    nah

  2. anonymous
    • one year ago
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    \[\frac{d^2y}{dx^2}+2\frac{dy}{dx}+y=0\] Let \[y(x)=e^{rx}\]\[\implies r^2e^{rx}+2re^{rx}+e^{rx}=0\]\[e^{rx}(r^2+2r+1)=0\] On solving the polynomial, \[r=-1\] therefore, \[y(x)=e^{-x}\]

  3. anonymous
    • one year ago
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    if \[y(x)=e^{rx}\] then \[\frac{dy}{dx}=re^{rx}\] and \[\frac{d^2y}{dx^2}=e^{rx}+r^2e^{rx}\] so we get: \[r^2e^{rx}+2re^{rx}+2e^{rx}=0\] the polynomial you will get is slightly different: \[r^2+2r+2=0\]

  4. anonymous
    • one year ago
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    why is that ??

  5. anonymous
    • one year ago
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    oh yeah ur right!!! let me try again

  6. anonymous
    • one year ago
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    noo wait the polynomial is correct

  7. anonymous
    • one year ago
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    I think your \[\frac{d^2y}{dx^2}\] is wrong

  8. IrishBoy123
    • one year ago
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    careful you have a **repeated root** !!!

  9. anonymous
    • one year ago
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    Oh sorry I forgot the constant \[y(x)=Ce^{-x}\]

  10. anonymous
    • one year ago
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    for the second derivative we need to use the chain rule, or derivative of a product.

  11. IrishBoy123
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx repeated roots

  12. anonymous
    • one year ago
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    yeah but r is constant so it's derivative would be 0 and u will only get 1 term

  13. anonymous
    • one year ago
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    ooppss!! haha! you are right! r is constant, my bad there!! sorry... take a look at the link @IrishBoy123 posted

  14. anonymous
    • one year ago
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    Yeh im checking it

  15. anonymous
    • one year ago
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    There is 1 more solution ??

  16. IrishBoy123
    • one year ago
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    the general solution for repeated roots is \( y = (Ax + B ) e^{ax} \)

  17. anonymous
    • one year ago
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    ok I read the article now so...the solution should be I think \[y(x)=C_{1}e^{-x}+C_{2}xe^{-x}\]

  18. IrishBoy123
    • one year ago
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    i don't know how you were shown how to to these, but with a differential operator D = d/dx, you take (D-a)(D-a)y = 0 and let u = (D-a)y, so (D-a)u = 0, giving \(u = A e^{ax}\) from there \( ( D-a)y = A e^{ax},\ \frac{dy}{dx} - ay =A e^{ax} \), solve with integrating factor

  19. IrishBoy123
    • one year ago
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    yes, Ax + B appears in general solution for repeated roots

  20. anonymous
    • one year ago
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    I only knew one method u take the polynomial and with the roots u do \[y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}\]

  21. IrishBoy123
    • one year ago
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    ok, so now also remember that, with repeated roots, that rule changes :p

  22. anonymous
    • one year ago
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    kk

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