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anonymous

  • 5 years ago

Use power series to solve y''(t)=y*y'(t), y(0)=2 and y'(0)=-3 I know to assume that y=a0+a1t+a2t^2+a3t^3+a4t^4 (where the number following a is its subscript and not a coefficient on t) and I know to get the first and second derivatives from there. What I am having trouble with is foiling the series for y and the series for y'. Could someone please help me finish this problem? If I could just multiply the two series together without any mistakes, I'd pretty much be done.

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  1. anonymous
    • 5 years ago
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    I answered your question earlier.

  2. anonymous
    • 5 years ago
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    thank you so much for that

  3. anonymous
    • 5 years ago
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    So it's okay?

  4. anonymous
    • 5 years ago
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    Would you be willing to help me foil out the two series? My teacher wants me to show them out to the fourth degree (something I forgot to write into my original question)

  5. anonymous
    • 5 years ago
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    the series for y and y' that is

  6. anonymous
    • 5 years ago
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    Do you mean your teacher wants you to expand to the fourth degree?

  7. anonymous
    • 5 years ago
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    yeah. I actually just did the foil for the fourth degree of y and y' . y*y'=a0a1+(2a0a2+(a1)^2)t+(3a0a3+3a1a2)t^2+(4a0a4+4a1a3+2(a2^2)t^3+(5a0a5+5a1a4+5a2a3)t^4. Here's my new problem though: Now that I have all of the values for them, I am told to equate different coefficients of powers of t, but I keep on screwing up the scratchwork. would you help me euate the different powers of t?

  8. Gina
    • 5 years ago
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    \[y \prime=10^{x+y}\]

  9. anonymous
    • 5 years ago
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    I'm not sure how your reply makes sense here Gina.

  10. anonymous
    • 5 years ago
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    That looks right. I haven't completed a total expansion, though. So, you're now needing to actually determine the coefficients? Are you allowed to use the formula I derived earlier?

  11. anonymous
    • 5 years ago
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    \[a_{n+2}=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}(n-k+1)a_na_{n-k+1}\]

  12. anonymous
    • 5 years ago
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    Yeah, but I think he just wants us to equate the different coefficients of different powers of t and solve algebraically. So for instance, the constant term for y''=yy' is 2a2=a0a1-->a2=(1/2)a0a1. then you go down the list and just put everything in terms of a0 and a1.

  13. anonymous
    • 5 years ago
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    Gina, post your question again separately and I'll help.

  14. anonymous
    • 5 years ago
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    does that make sense?

  15. anonymous
    • 5 years ago
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    Yep. You have a0 and a1...the rest is just turgid algebra. I get the feeling you're screwing it up because it's tedious and has many 'moving parts'. It sounds like you know what to do, though. I don't know what else I can do.

  16. anonymous
    • 5 years ago
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    Yeah, it sounds like your teacher just wants you to go through the motions.

  17. anonymous
    • 5 years ago
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    help me with the actual algebra part? I really need to get this question done and all that I'm screwing up is the tedious part.

  18. anonymous
    • 5 years ago
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    I'll see if I can get you started.

  19. anonymous
    • 5 years ago
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    thank you so much. I think I can get the first few t terms without trouble. It's like the t^3 and t^4 terms taht are giving me trouble.

  20. anonymous
    • 5 years ago
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    any luck?

  21. anonymous
    • 5 years ago
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    I expanded first to check, which took some time.

  22. anonymous
    • 5 years ago
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    \[2a_2=a_0a_1\rightarrow a_2=\frac{2 \times -3}{2}=-3\] \[6a_3=2a_0a_2+a^2_1\rightarrow a_3=\frac{2 \times 2 \times -3+(-3)^2}{6}=-\frac{1}{2}\] Is this looking familiar?

  23. anonymous
    • 5 years ago
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    \[12a_4=3a_0a_3+3a_1a_2\]\[\rightarrow a_4=\frac{3*2*-\frac{1}{2}+3*-3*-3*-3}{12}=-7\]

  24. anonymous
    • 5 years ago
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    \[20a_5=4a_0a_4+4a_1a_3+2a^2_2\rightarrow a_5=-\frac{8}{5}\]

  25. anonymous
    • 5 years ago
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    \[30a_6=5a_0a_5+5a_1a_4+5a_2a_3 \rightarrow a_6=\frac{193}{60}\]

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