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goalie2012

  • 2 years ago

solve the differential equation with initial conditions; compute the first 6 coefficients(a0-a5); find the general pattern: (1-2x)y''-y'+xy=0 y(0)=0, y'(0)=1

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  1. oldrin.bataku
    • 2 years ago
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    Have you read the FAQ? It's generally frowned upon to post the same question multiple times... additionally, you are expected to reward helpful answers with medals (re: your previous question)

  2. goalie2012
    • 2 years ago
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    sorry about that. Kind of new. thanks for the tip

  3. oldrin.bataku
    • 2 years ago
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    Anyways, again consider an analytic solution \(y=\sum\limits_{n=0}^\infty a_nx^n\) with derivatives $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n$$

  4. oldrin.bataku
    • 2 years ago
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    Plugging into our equation we have:$$(1-2x)\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty(n+1)a_{n+1}x^n+x\sum_{n=0}^\infty a_nx^n=0$$

  5. oldrin.bataku
    • 2 years ago
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    For our last term, note we can multiply into our sum \(x\) and shift our indices of summation:$$x\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1}=\sum_{n=1}^\infty a_{n-1}x^n$$

  6. oldrin.bataku
    • 2 years ago
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    For our first term, observe we can distribute \(1-2x\) and do something similar:$$\begin{align*}(1-2x)&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-2x\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2(n+1)(n+2)a_{n+2}x^{n+1}\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\end{align*}$$

  7. oldrin.bataku
    • 2 years ago
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    @goalie2012 you should award the medal on the other question... I haven't answered this one yet!

  8. oldrin.bataku
    • 2 years ago
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    Now, I haven't combined them into 1 yet because dealing with the starting indices would be gross...

  9. goalie2012
    • 2 years ago
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    ya. that's where I got a little lost

  10. oldrin.bataku
    • 2 years ago
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    Well, we can safely rewrite that summation again with 'nicer' indices since starting at \(n=0\) will just add \(2(0)(1)a_1x^0=0\):$$\begin{align*}&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n\end{align*}$$

  11. oldrin.bataku
    • 2 years ago
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    We can also extend our previous summation:$$\sum_{n=1}^\infty a_{n-1}x^n=\sum_{n=0}^\infty a_{n-1}x^n$$where we can impose \(a_m=0\) for \(m<0\) (this way we don't change the value of the summation); this is okay because any \(a_m\) for \(m<0\) do not appear in our solution.

  12. oldrin.bataku
    • 2 years ago
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    Anyways, let's put it all together:$$\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n-\sum_{n=0}^\infty (n+1)a_{n+1}x^n+\sum_{n=0}^\infty a_{n-1}x^n=0\\\sum_{n=0}^\infty\left(\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}\right)x^n=0$$Because our right must be identically \(0\), we require the coefficient of each power of \(x\) to be \(0\):$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$

  13. oldrin.bataku
    • 2 years ago
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    Now, consider our initial conditions \(y(0)=0,y'(0)=1\):$$y(0)=0\\\sum_{n=0}^\infty a_nx^n=0$$Since \(x^n=0\) for \(n>0\), we're left with only the term for \(n=0\):$$a_0x^0=0\\a_0=0$$

  14. oldrin.bataku
    • 2 years ago
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    Moving on to our second condition \(y'(0)=1\), we find:$$\sum_{n=0}^\infty(n+1)a_{n+1}x^n=1$$Again, \(x=0\) leaves all but our \(n=0\) term:$$a_1x^0=1\\a_1=1$$... so we've solved for first two coefficients, \(a_0=0,a_1=1\).

  15. oldrin.bataku
    • 2 years ago
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    For our last three, we go back to that expression!

  16. oldrin.bataku
    • 2 years ago
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    $$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$Letting \(n=0\) we know \(a_{-1}=0,a_0=0,a_1=1\) so:$$(2a_2-a_1)+a_{-1}=0\\2a_2-1=0\\a_2=\frac12$$

  17. oldrin.bataku
    • 2 years ago
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    For \(n=1\) we find:$$2(3a_3-2a_2-a_2)+a_0=0\\6a_3-6a_2+a_0=0$$Knowing \(a_0=0,a_1=1\) and having just concluded \(a_2=\frac12\):$$6a_3-3=0\\6a_3=3\\a_3=\frac12$$

  18. oldrin.bataku
    • 2 years ago
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    $$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$ For \(n=2\) we find:$$3(4a_4-2(2)a_3-a_3)+a_1=0\\12a_4-15a_3+a_1=0$$Again, knowing \(a_1=1\) and having concluded \(a_3=\frac12\):$$12a_4-\frac{15}2+1=0\\24a_4-15+2=0\\24a_4-13=0\\a_4=\frac{13}{24}$$

  19. oldrin.bataku
    • 2 years ago
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    Can you take it from here? ;-)

  20. goalie2012
    • 2 years ago
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    yep. thank you very much. I have two more. would you be willing to help?

  21. oldrin.bataku
    • 2 years ago
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    I can try!

  22. goalie2012
    • 2 years ago
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    one won't take long and the other I'm not sure about

  23. goalie2012
    • 2 years ago
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    I have to close this question to write another one, right?

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