Euler's method for exact solution We know that the solution of
$$y'=y$$
with $y(0)=1$
is 
$$y=e^{x}$$
As far as I see the Euler's method is explicitly used only to find the numerical approximation of e.g. $y(3)$. Can we use the Euler's method to solve this differential equation and find the exact solution?
Edit:
I thought of the following:
$$y(t_1+\delta t) = y(t_0) + y'(t_1) \delta t$$
$$y(t_1+2\delta t) = y(t_1+\delta t) + y'(t_1+\delta t) \delta t$$
$$...$$
and from that we could find a recursive solution for $y(t_0)$. Then by taking the limit $\lim_{\delta t \to 0}$ of it, it could be possible to find the general solution of $y$ for every $t_0$ in the domain.
Is it impossible?
 A: I think that the process you're envisioning, if you did it rigorously, becomes equivalent to Picard's iterative process: Given an initial value problem $y'(t)= f(t,y)$, $y(0) = y_0$, the sequence of functions defined by
\begin{align*}
    \phi_0(t) &= y_0 \\
    \phi_{k+1}(t) &= y_0 + \int_0^t f(s,\phi_k(s))\,ds
\end{align*}
converges to a solution $\phi(t)$.
(You can think of that integral as summing $f(s,\phi_k(s))$ at many discrete points between $s=0$ and $s=t$, multiplied by a small amount $\Delta s$, and taking the limit as $\Delta s \to 0$.)
In your problem, you have $f(t,y) = y$ and $y_0 = 1$.  Picard's process gives:
\begin{align*}
    \phi_0(t) &= 1 \\
    \phi_1(t) &= 1 + \int_0^t \phi_0(s)\,ds = 1 + \int_0^t 1\,ds = 1 + t \\
    \phi_2(t) &= 1 + \int_0^t(1+s)\,ds = 1 + t + \tfrac{1}{2}t^2 \\
    \phi_3(t) &= 1 + \int_0^t\left(1 + s + \tfrac{1}{2}s^2\right)\,ds = 1 + t + \tfrac{1}{2}t^2 + \tfrac{1}{6} t^3
\end{align*}
You can see that $\phi_k(t)$ is the $k$-th degree Taylor polynomial for $\phi(t) = e^t$, and that is the solution to the IVP.
A: No.  Euler's method is only an approximation.  To determine the exact value of $y$ at time $t+\delta t$ (regardless of whether the ODE has an exact solution), you would need to keep all terms of the Taylor expansion for the solution.  Euler's method gives
$$y(t+\delta t) \approx y(t) + y'(t) \delta t,$$
whereas the exact solution at $t+\delta t$ is
$$y(t+\delta t) = y(t) + y'(t) \delta t + \frac{1}{2}y''(t)\delta t^2 + \cdots.$$
In particular, Euler's method will only be exact if the solution is affine (of the form $y = ax+b$) so that all derivatives beyond the first derivative are zero.
