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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
 10 months ago
 10 months 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
 10 months ago
 10 months ago

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oldrin.batakuBest ResponseYou've already chosen the best response.1
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)
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
sorry about that. Kind of new. thanks for the tip
 10 months ago

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

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

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

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

oldrin.batakuBest 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!
 10 months ago

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

goalie2012Best ResponseYou've already chosen the best response.0
ya. that's where I got a little lost
 10 months ago

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

oldrin.batakuBest ResponseYou've already chosen the best response.1
We 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.
 10 months ago

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

oldrin.batakuBest ResponseYou've already chosen the best response.1
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 \(n=0\):$$a_0x^0=0\\a_0=0$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
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, \(a_0=0,a_1=1\).
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
For our last three, we go back to that expression!
 10 months ago

oldrin.batakuBest 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:$$(2a_2a_1)+a_{1}=0\\2a_21=0\\a_2=\frac12$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
For \(n=1\) 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 \(a_2=\frac12\):$$6a_33=0\\6a_3=3\\a_3=\frac12$$
 10 months ago

oldrin.batakuBest 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 \(a_3=\frac12\):$$12a_4\frac{15}2+1=0\\24a_415+2=0\\24a_413=0\\a_4=\frac{13}{24}$$
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.1
Can you take it from here? ;)
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
yep. thank you very much. I have two more. would you be willing to help?
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
one won't take long and the other I'm not sure about
 10 months ago

goalie2012Best ResponseYou've already chosen the best response.0
I have to close this question to write another one, right?
 10 months ago
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