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Solve $y'=y$, $y(0)=1$ using method of successive approximations, obtaining the power series expansions of the solution.


Solving the above initial value problem is equivalent to solving (for $\psi$) the integral equation:

$$\psi(x)=\int_0^x \psi(t)dt + 1 \mbox{.}$$

Consider a continuous map $F: C(\mathbb{R})\rightarrow C(\mathbb{R})$ defined by

$$(F(\psi))(x)=\int_0^x \psi(t)dt +1 \mbox{.}$$

I have already proved that such $F$ is a contraction (it's easy in this case). Thus there exist a unique fixed point $\phi$.

Start the iterative procedure with $\psi_0=0$.

$$(F(\psi_0))(x)=1$$

$$\psi_1=1$$

$$(F(\psi_1))(x)=x+1$$

$$\psi_2=x+1$$

$$(F(\psi_2))(x)=\frac{x^2}{2}+x+1$$

$$\psi_3=\frac{x^2}{2}+x+1$$

$$(F(\psi_3))(x)=\frac{x^3}{6}+\frac{x^2}{2}+x+1$$

$$\ldots$$

As we suspected, it looks similiar to the expansion of $y(x)=e^x$ in Taylor series. Using induction we can prove that it is indeed the solution.

Honestly speaking, I have found this method of finding solutions to ODEs in some old book. I have never done this before (using the contraction theorem). I would be happy is someone verified this.

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Your work is correct (assuming that your proof of contractiveness is correct). This is part of a general theory which includes the Picard-Lindelöf theorem, and using fixed-point theorems is a very standard way of proving existence results.

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  • $\begingroup$ I have wondered a bit about my solution and I think there is a slight mistake. I cannot use Banach's fixed point theorem on such $F$ because $C(\mathbb{R})$ is not complete. But I can use it on any $C([-r,r])$ with arbitrarly big $r$. It this a good justification? $\endgroup$ – luka5z Dec 20 '15 at 8:16

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