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 one year ago
solve the differential equation with initial conditions; compute the first 6 coefficients(a0a5); find the general pattern:
(12x)y''y'+xy=0 y(0)=0, y'(0)=1
 one year ago
solve the differential equation with initial conditions; compute the first 6 coefficients(a0a5); find the general pattern: (12x)y''y'+xy=0 y(0)=0, y'(0)=1

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oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Have 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)

goalie2012
 one year ago
Best ResponseYou've already chosen the best response.0sorry about that. Kind of new. thanks for the tip

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Anyways, again consider an analytic solution \(y=\sum\limits_{n=0}^\infty a_nx^n\) with derivatives $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Plugging into our equation we have:$$(12x)\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\sum_{n=0}^\infty(n+1)a_{n+1}x^n+x\sum_{n=0}^\infty a_nx^n=0$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1For our last term, note we can multiply into our sum \(x\) and shift our indices of summation:$$x\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1}=\sum_{n=1}^\infty a_{n1}x^n$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1For our first term, observe we can distribute \(12x\) and do something similar:$$\begin{align*}(12x)&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n2x\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\sum_{n=0}^\infty2(n+1)(n+2)a_{n+2}x^{n+1}\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\end{align*}$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1@goalie2012 you should award the medal on the other question... I haven't answered this one yet!

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Now, I haven't combined them into 1 yet because dealing with the starting indices would be gross...

goalie2012
 one year ago
Best ResponseYou've already chosen the best response.0ya. that's where I got a little lost

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Well, we can safely rewrite that summation again with 'nicer' indices since starting at \(n=0\) will just add \(2(0)(1)a_1x^0=0\):$$\begin{align*}&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\sum_{n=0}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty[(n+2)a_{n+2}2na_{n+1}](n+1)x^n\end{align*}$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1We can also extend our previous summation:$$\sum_{n=1}^\infty a_{n1}x^n=\sum_{n=0}^\infty a_{n1}x^n$$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.

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Anyways, let's put it all together:$$\sum_{n=0}^\infty[(n+2)a_{n+2}2na_{n+1}](n+1)x^n\sum_{n=0}^\infty (n+1)a_{n+1}x^n+\sum_{n=0}^\infty a_{n1}x^n=0\\\sum_{n=0}^\infty\left(\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}\right)x^n=0$$Because our right must be identically \(0\), we require the coefficient of each power of \(x\) to be \(0\):$$\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}=0$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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 \(n=0\):$$a_0x^0=0\\a_0=0$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Moving 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, \(a_0=0,a_1=1\).

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1For our last three, we go back to that expression!

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1$$\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}=0$$Letting \(n=0\) we know \(a_{1}=0,a_0=0,a_1=1\) so:$$(2a_2a_1)+a_{1}=0\\2a_21=0\\a_2=\frac12$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1For \(n=1\) we find:$$2(3a_32a_2a_2)+a_0=0\\6a_36a_2+a_0=0$$Knowing \(a_0=0,a_1=1\) and having just concluded \(a_2=\frac12\):$$6a_33=0\\6a_3=3\\a_3=\frac12$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1$$\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}=0$$ For \(n=2\) we find:$$3(4a_42(2)a_3a_3)+a_1=0\\12a_415a_3+a_1=0$$Again, knowing \(a_1=1\) and having concluded \(a_3=\frac12\):$$12a_4\frac{15}2+1=0\\24a_415+2=0\\24a_413=0\\a_4=\frac{13}{24}$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.1Can you take it from here? ;)

goalie2012
 one year ago
Best ResponseYou've already chosen the best response.0yep. thank you very much. I have two more. would you be willing to help?

goalie2012
 one year ago
Best ResponseYou've already chosen the best response.0one won't take long and the other I'm not sure about

goalie2012
 one year ago
Best ResponseYou've already chosen the best response.0I have to close this question to write another one, right?
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