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I have to solve this non-linear DE

$y' -e^y -x^2 = 0 , y(0)=c$

using powerseries.

$y(x) = \sum_{n=0}^\infty a_{n}x^n $

$y'(x) = \sum_{n=1}^\infty na_{n}x^{n-1} $

so we get

$\sum_{n=1}^\infty na_{n}x^{n-1} -e^{\sum_{n=0}^\infty a_{n}x^n} -x^2 = 0 $

changing the sum indexes

$\sum_{n=0}^\infty (n+1)a_{n+1}x^{n} -e^{\sum_{n=0}^\infty a_{n}x^n} -x^2 = 0 $

I want to find the recurrence relation of the coefficients. But i cannot merge/group them into the same sum, meaning, the sum of $y'(x) = \sum_{n=1}^\infty na_{n}x^{n-1} $ and the sum of $y(x) = \sum_{n=0}^\infty a_{n}x^n $ cause it is power of $e$ in order to get the equality of them with $0$.

DontWorry
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  • It would probably be better to let $y = \ln z$. – Matthew Cassell May 22 '22 at 17:10
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    Are you required to find a power series for $y$ or just a solution using power series? With $u=e^{-y}$ you get the equation $u'=-e^{-y}y'=-1-x^2u$ which is now linear and thus easier to solve for a power series for $u$. – Lutz Lehmann May 22 '22 at 19:42

2 Answers2

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The power series for $e^y$ can be computed via the differential equation of the anti-logarithm. Wtih $v=e^{y}=\sum b_nx^n$ one has $$ v'=vy'\implies \sum nb_nx^{n-1}=\sum b_m\,(n-m)a_{n-m}x^{n-1} \\~\\ b_0=e^{a_0},\\ b_n=\frac1n\sum_{m=0}^{n-1}(n-m)b_ma_{n-m}. $$ This allows the recursive computation of the coefficients $b_n$.

Lutz Lehmann
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  • Yes, thank you, corrected. – Lutz Lehmann May 22 '22 at 22:05
  • I am a bit lost why did you set $u=e^y$ and not $u=e^{-y}$ like you proposed in the comment of my question? Multiplying by $e^{-y}$ we manage to group $y'$ with $e^{-y}$ creating the $u'$ so we end up with $u' = -1-x^2*u$, and now we got 2 coefficients. Also i did not quite understand how you got $\sum b_m,(n-m)a_{n-m}x^{n-1}.$ – DontWorry May 22 '22 at 22:45
  • Because that is required if you want to compute the power series of $y$ directly. // This results from the power series multiplication $vy'$ with terms $b_mx^m$ and $ka_kx^{k-1}$, combining the products with the same resulting degree $m+(k-1)=(n-1)$ or $k=n-m$. – Lutz Lehmann May 23 '22 at 05:16
  • Because this is the first time I have to solve a non linear DE this way and I still have many questionmarks, can you provide me with a link that might help? solved problems? methodology? etc. – DontWorry May 23 '22 at 13:54
  • This technique is called Taylor series arithmetic. It is one of the main methods of automatic differentiation.You can find it in books on that topic, by Griewank or Naumann. There also exists a C++ template library TADIFF as part of FADBAD (older name), the documentation has papers that also explain the basics. An example with more details is in https://math.stackexchange.com/questions/2432062/how-solve – Lutz Lehmann May 23 '22 at 14:32
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The differential equation to be solved is $$ y' -e^y -x^2 = 0 ,\qquad y(0)=c $$ and the suggested approach is with power series. It may not be immediately obvious, but after working with the power series for $\,y\,$ a good step is to define $\, u := e^{-y}.\,$ Rewrite$\,y\,$ in terms of $\,u\,$ in the D.E. to get $$ -\log(u)' - 1/u - x^2 = 0, \qquad u(0) = c_0 := e^{-c}. $$ Simplify to get a new differential equation $$ u' = -1 - u\,x^2. $$ Expand $\,u\,$ as a power series to get $$ u \!=\! \sum_{n=0}^\infty c_n \frac{x^n}{n!}, u' \!=\! \sum_{n=0}^\infty c_{n+1} \frac{x^n}{n!}, u\,x^2 \!=\! \sum_{n=0}^\infty c_{n-2}n(n\!-\!1) \frac{x^n}{n!}. $$ Substitute these series into the new D.E. to get $$ c_1 = -1,\quad c_{n+1} = -c_{n-2}n(n-1) \quad\text{ unless } \quad n=0. $$ Use this recursion relation to get $$ u = c_0-x-2c_0\frac{x^3}{3!} +6\frac{x^4}{4!} +40c_0\frac{x^6}{6!}-180\frac{x^7}{7!}+\cdots.$$ Use $\,y = -\log(u/c_0)+c\,$ to get the solution $$ c + \frac1{c_0}\frac{x^1}{1!} + \frac1{c_0^2} \frac{x^2}{2!} + 2\frac{1+c_0^3}{c_0^3}\frac{x^3}{3!} + 2\frac{3+c_0^3}{c_0^4}\frac{x^4}{4!} + \cdots. $$

Somos
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