## The method

Suppose we have an ODE (with initial conditions) $x'\!\left(t\right)=f\!\left(x\!\left(t\right),t\right), \quad x\!\left(t_0\right)=C,$$\p{1}$ where $x$ is the unknown function, and $f$ is Lipschitz continuous in its first argument and continuous in its second argument. By Picard–Lindelöf theorem, we can seek the unique solution within $t\in\left[t_0-\varepsilon,t_0+\varepsilon\right]$.

Here, I propose the following method: we can write out a sequence of functions defined by $x_0\!\left(t\right)\coloneqq C,$$\p{2}$ $x_{n+1}\!\left(t\right)\coloneqq\int_{t_0}^tf\!\left(x_n\!\left(s\right),s\right)\,\mathrm ds+C$$\p{3}$ (the properties of $f$ guarantee that the integral is well-defined). Then, if the sequence $\left(x_n'\right)$ converges uniformly on $\left[t-\varepsilon,t+\varepsilon\right]$ (question: can this condition actually be proved?), then the sequence $\left(x_n\right)$ converges uniformly to a function $x$ on $\left[t-\varepsilon,t+\varepsilon\right]$, which is the unique solution to Equation 1.

## Proof

The proof is easy. Note from Equation 3 that $x_{n+1}'\!\left(t\right)=f\!\left(x_n\!\left(t\right),t\right).$ Then, take the limit $n\to\infty$. By the uniform convergence, we recovers Equation 1.

## An example

Suppose $f\!\left(x,t\right)\coloneqq x$ and $t_0\coloneqq0$, then $x_n\!\left(t\right)=C\sum_{j=0}^{n-1}\frac{t^j}{j!}.$ This is because $\int_0^tx_n\!\left(s\right)\,\mathrm ds+C=C\sum_{j=0}^{n-1}\frac{t^{j+1}}{j!\left(j+1\right)}+C=x_{n+1}\!\left(t\right).$ By taking the limit $n\to\infty$, we get $x\!\left(t\right)=C\mathrm e^t$, which means that the unique solution to $x'\!\left(t\right)=x\!\left(t\right)$ with initial condition $x\!\left(0\right)=C$ is $x\!\left(t\right)=C\mathrm e^t$, expectedly.

## Approximation

This method is good because integration is sometimes much easier than solving ODE. We can use integration to get functions that are close to the exact solution. A natural question to ask is how close $x_n$ is to the exact solution $x$.

If Equation 1 is defined on $\mathbb R^n$ (where we may define differences and infinitesimals), it is clear that $x_n-x$ is an infinitesimal of higher order than $\left(t-t_0\right)^n$.