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anonymous
 3 years 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
anonymous
 3 years 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Have 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry about that. Kind of new. thanks for the tip

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@goalie2012 you should award the medal on the other question... I haven't answered this one yet!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, I haven't combined them into 1 yet because dealing with the starting indices would be gross...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya. that's where I got a little lost

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyways, 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 \(0\):$$\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}=0$$

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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 \(n=0\):$$a_0x^0=0\\a_0=0$$

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Moving 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, \(a_0=0,a_1=1\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For our last three, we go back to that expression!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For \(n=1\) 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 \(a_2=\frac12\):$$6a_33=0\\6a_3=3\\a_3=\frac12$$

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0$$\left[(n+2)a_{n+2}2na_{n+1}a_{n+1}\right](n+1)+a_{n1}=0$$ 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 \(a_3=\frac12\):$$12a_4\frac{15}2+1=0\\24a_415+2=0\\24a_413=0\\a_4=\frac{13}{24}$$

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you take it from here? ;)

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

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

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