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The ODE I have is $$y'(x)+e^{y(x)}+\frac{e^x-e^{-x}}{4}=0, \hspace{0.2cm} y(0)=0$$ I want to determine the first five terms (coefficients $a_0,\ldots, a_5$) of the power series solution $$y(x)=\sum_{k=0}^{\infty} a_kx^k$$ So far, I know that $$y'(x)=\sum_{k=1}^{\infty} a_kkx^{k-1}$$ Now I plug these back into the equation and get: $$\sum_{k=1}^{\infty} a_kkx^{k-1} + e^{\sum_{k=0}^{\infty} a_kx^k} + \frac{e^x-e^{-x}}{4}=0$$. Now I'm not sure how to continue with this. Please help.

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    $\begingroup$ I didn't know Lebron was studying ODE, but you might have a better time by expanding $e^y$ in terms of its power series, then plug in the power series approximation for $y$ from there. $\endgroup$ – DaveNine Mar 27 '16 at 22:08
  • $\begingroup$ Could you elaborate on that a bit? $\endgroup$ – LeBron James Mar 28 '16 at 2:45
  • $\begingroup$ $e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!}$, use this to make computing the coefficients easier. I'd consider truncating the series in order to compute the coefficients you want as well. $\endgroup$ – DaveNine Mar 28 '16 at 3:35
  • $\begingroup$ I used the expansion for $e^x$ and $e^{-x}$, but I fail to see how this will make $e^y$ easier, now I have $$e^y = \sum_{k=0}^{\infty}\frac{\sum_{k=0}^{\infty} a_kx^k}{k!}$$ $\endgroup$ – LeBron James Mar 28 '16 at 3:55
  • $\begingroup$ Another hint, you're going to want to move over the taylor series of $\frac{1}{2} \cosh(x)$ term to the other side, and then match coefficients. This should give you a means to solve for coefficients. $\endgroup$ – DaveNine Mar 28 '16 at 3:56
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Since you just need a few terms, setting $$y=\sum_{n=1}^6 a_i x^i$$ (because of the condition $y(0)=0$) you could develop $e^y$ as a Taylor series around $x=0$ and get $$e^y=1+a_1 x+\frac{1}{2} \left(a_1^2+2 a_2\right) x^2+\frac{1}{6} \left(a_1^3+6 a_2 a_1+6 a_3\right) x^3+$$ $$\frac{1}{24} \left(a_1^4+12 a_2 a_1^2+24 a_3 a_1+12 a_2^2+24 a_4\right) x^4+$$ $$\frac{1}{120} \left(a_1^5+20 a_2 a_1^3+60 a_3 a_1^2+60 a_2^2 a_1+120 a_4 a_1+120 a_2 a_3+120 a_5\right) x^5+$$ $$\frac{1}{720} \left(a_1^6+30 a_2 a_1^4+120 a_3 a_1^3+180 a_2^2 a_1^2+360 a_4 a_1^2+720 a_2 a_3 a_1+720 a_5 a_1+120 a_2^3+360 a_3^2+720 a_2 a_4+720 a_6\right) x^6$$

Expanding the term $\frac{e^x-e^{-x}}{4}$ as a Taylor series too, the differential equation then write $$(a_1+1)+\left(a_1+2 a_2+\frac{1}{2}\right) x+\frac{1}{2} \left(a_1^2+2 a_2+6 a_3\right) x^2+\frac{1}{12} \left(2 a_1^3+12 a_2 a_1+12 a_3+48 a_4+1\right) x^3+$$ $$\frac{1}{24} \left(a_1^4+12 a_2 a_1^2+24 a_3 a_1+12 a_2^2+24 a_4+120 a_5\right) x^4+$$ $$\frac{1}{240} \left(2 a_1^5+40 a_2 a_1^3+120 a_3 a_1^2+120 a_2^2 a_1+240 a_4 a_1+240 a_2 a_3+240 a_5+1440 a_6+1\right) x^5+$$ $$\frac{1}{720} \left(a_1^6+30 a_2 a_1^4+120 a_3 a_1^3+180 a_2^2 a_1^2+360 a_4 a_1^2+720 a_2 a_3 a_1+720 a_5 a_1+120 a_2^3+360 a_3^2+720 a_2 a_4+720 a_6\right) x^6=0$$

Cancelling the coefficients lead to $$a_1=-1\qquad a_2=\frac{1}{4}\qquad a_3=-\frac{1}{4}\qquad a_4=\frac{7}{48}\qquad a_5=-\frac{19}{160}$$

I hope I did not make any mistake since my results do not coincide with DaveNine's answer.

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  • $\begingroup$ The issue with this is that, through $e^y$, each coefficient $a_1,a_2,...$ are expressed as an infinite series themselves. This means your answers are more of an approximation rather than exact. $\endgroup$ – DaveNine Mar 28 '16 at 5:50
  • $\begingroup$ @DaveNine. I totally agree with you ! I must confess that I was not very comfortable after writing. Your approach is definitely more rigorous. What I do not understand is why, even for the very first terms, I do not get the same coefficients as yours. Cheers. $\endgroup$ – Claude Leibovici Mar 28 '16 at 5:55
  • $\begingroup$ Ah I see. Well each coefficient can be represented as a power series because of the exponential term, that's probably why. If you truncate that series then you approximate every coefficient no matter the location in the series of $y$. $\endgroup$ – DaveNine Mar 28 '16 at 6:07
  • $\begingroup$ thanks, I got the same answers but using a repetitive derivative method, basically solving for $y'(0), y'(0), y''(0)...etc$, then using the Taylor formula to get the coefficients. $\endgroup$ – LeBron James Mar 28 '16 at 6:17
  • $\begingroup$ I was incorrect, I expanded the non-derivative term incorrectly (You don't even need to anyways.) $\endgroup$ – DaveNine Mar 28 '16 at 8:05
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The answers provided are excellent, but I'll offer what I think is the easiest solution: Just take derivatives of the equation, plug in $0$ and get as many $y^{(n)}(0)$'s as you want, then finally construct your taylor series:

$$y=\sum_{n=0}^\infty \frac{y^{(n)}(0)}{n!}x^n.$$

$$y'+e^{y}+\frac{1}{2}\sinh(x)=0, \quad\quad y(0)=0$$ Plugging in $y(0)=0$, we get $y'(0)+1=0$, thus $\boxed{y'(0)=-1}$.

Now differentiate the original equation as many times as needed to get the number of coefficients that you want: $$y''+y'e^y+\frac{1}{2}\cosh(x)=0$$ $$y'''+(y')^2e^y+y''e^y+\frac{1}{2}\sinh(x)=0$$ etc...

Plugging in $x=0$ into the above equations gives: $$y''(0)+y'(0)e^{y(0)}+\frac{1}{2}\cosh(0)=0 \quad \Rightarrow \quad \boxed{y''(0)=\frac{1}{2}}$$ $$y'''(0)+(y'(0))^2e^y+y''(0)e^{y(0)}+\frac{1}{2}\sinh(0)=0 \quad \Rightarrow \quad \boxed{y'''(0)=-\frac{3}{2}}$$

Thus: $$ \begin{aligned} y&=y(0)+y'(0)x+\frac{1}{2!}y''(0)x^2+\frac{1}{3!}y'''(0)x^3+\cdots \\ &=-x+\frac{1}{2}\frac{1}{2!}x^2-\frac{3}{2}\frac{1}{3!}x^3+\cdots \\ &=-x+\frac{1}{4}x^2-\frac{1}{4}x^3+\cdots \end{aligned} $$

This method is nice because you don't have to work out recursion relations, etc.

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To elaborate on my hint (it my not be the best method, but it should work.)

We have

$$\sum_{k=1}^{\infty} ka_{k}x^{k-1} + \exp \left( \sum_{n=0}^{\infty} a_n x^n \right) = \frac{1}{4}(\exp{(x)} - \exp{(-x)}) $$

Setting $x = 0$ and using $a_0 = 0$ we've

$$1+a_1 = 0$$

which gives $a_1 = -1$.

To get other coefficients, one can simply differentiate in the above equation with respect to $x$, then set $x = 0$ to get rid of any other terms not used. Doing this, I get $a_2 = \frac{1}{4}$, $a_3 = \frac{-1}{4}$, $a_4 = \frac{7}{48}$, and $a_5 = \frac{-19}{160}$.

My apologies to Claude!

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What we are after is just some $k$'th order Taylor expansion of $y(x)$ about $x=0$. From Taylor's theorem we have that the coefficients $a_n$ in $$f(x) = \sum_{n=0}^\infty a_n x^n$$ satisfy $a_n = \frac{f^{(n)}(0)}{n!}$. We can compute this recursively by taking the $n$'th derivative of the ODE, $y'(x) = -e^{y(x)} - \frac{\sinh(x)}{2}$, to get $$y^{(n+1)}(0) = \left.\frac{d^n}{dx^n}e^{y(x)}\right|_{x=0} - \frac{1}{2}\left.\frac{d^n\sinh(x)}{dx^n}\right|_{x=0}$$ Using the expression for the $n$'th derivative of $e^{y(x)}$ we obtain the recursion

$$a_{n+1} = -\frac{1}{n+1}\left(\sum_{(m_1,\ldots,m_n)} \prod_{j=1}^n\frac{a_j^{m_j}}{m_j!}\right) -\frac{1-(-1)^n}{4(n+1)!}$$

where the sum extends over all $m$-tuples $(m_1,m_2,\ldots,m_n)$ of non-negative integers satisfying $m_1+2m_2+\ldots+nm_n = n$.

The formula above is not very useful when computing it by hand but it can be useful on a computer to automatize the computation of the coefficients. A simple (far from optimal) Mathematica implementation of this is given below and gives the following coefficients for $1\leq i \leq 20$:

$$a_i = \left\{-1,\frac{1}{4},-\frac{1}{4},\frac{7}{48},-\frac{19}{160},\frac{127}{1440},-\frac{185}{2688},\frac{125}{2304},-\frac{84131}{1935360},\frac{255841}{7257600},-\frac{12293681}{425779200},\frac{3263593}{136857600},-\frac{526926271}{26568622080},\frac{2893020049}{174356582400},-\frac{22242690527}{1594117324800},\frac{493179580879}{41845579776000},-\frac{43334418279277}{4335999123456000},\frac{8376548037343}{984980570112000},-\frac{150701255607005953}{20760763803107328000},\frac{465109942159859}{74858523328512000}\right\}$$

nmax = 20;
aa = Table[-1, {i, 1, nmax}];
varlist = Table[Subscript[x, i], {i, 1, nmax}];
Do[
  (* Variables *)
  var = varlist[[1 ;; i]];

  (* Define constraint on tuples *)
  constraint = Sum[j varlist[[j]], {j, 1, i}] == i;
  Do[constraint = constraint && var[[j]] >= 0, {j, 1, i}];

  (* Compute allowed tuples *)    
  tuples = var /. Solve[constraint, var, Integers];

  (* Compute sum over tuples *)
  sum = 0;
  Do[
   curtuple = tuples[[j]];
   (* Product over elements in tuples*)
   prod = 1;
   Do[prod *= 1/curtuple[[k]]! aa[[k]]^curtuple[[k]]];, {k, 1, i}];
   sum += prod;
   , {j, 1, Length[tuples]}];

   (* Store coefficient a(i+1) *)
  aa[[i + 1]] = -(sum/(i + 1)) - (1 - (-1)^i)/(4 (i + 1)!);
  Print[i + 1, "   ", aa[[i + 1]]];
  , {i, 1, nmax - 1}];
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