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Sometimes it is just not possible to solve an 1st order IVP \(y'=f(x,y)\), \(y(x_{0})=y_{0}\) analytically for an
implicit or explicit solution.
But you can estimate future values of the solution. The following method is
one way to do this, called Euler’s method (also called the \(``\)forward Euler"
method):
Suppose we are interested in finding the values of the solution \(y\) at equally spaced
values \(x_{i}=x_{0}+i\Delta x\) for \(i=1,2,...,n\) where \(\Delta x\) is the \(``\)step size."
Denote by \(y_{i}\) the approximate value \(y(x_{i})\) (for \(i=1,2,..,n\)). Then the absolute error at the ith step is \(e_{i}=|y(x_{i})-y_{i}|\)
(for \(i=1,2,..,n\)).
Euler’s method rests on having the tangent line to the graph of the solution curve
at \(x_{i}\) approximate the curve on \([x_{i},x_{i+1}]\). The slope at \((x_{i},y_{i})\) is \(f(x_{i},y_{i})\) so the tangent line on \([x_{i},x_{i+1}]\)
is
\[y=y_{i}+f(x_{i},y_{i})(x-x_{i}).\]
So for \(i=0\), we get the approximate value \(y(x_1)\approx y_{1}=y_{0}+f(x_{0},y_{0})(x_{1}-x_{0})=y_{0}+ f(x_{0},y_{0})\Delta x\).
Then we can repeat to get the approximate value \(y(x_2)\approx y_{2}=y_{1}+ f(x_{1},y_{1})\Delta x\).
We continue in this way.
At the \(i^{\mbox {th}}\) step it looks like \(y(x_i)\approx y_{i}=y_{i-1}+ f(x_{i-1},y_{i-1})\Delta x\) (for \(i=1,2,...,n\)).
Let’s use Euler’s method to estimate \(y(1)\) where \(y\) is the solution \(y'=x-y^2\), \(y(0)=1\) with a step size of
\(\Delta x=0.1\).
A direction field.
Figure 1: A direction field
Let \(f(x,y)=x-y^2\), \(y_0=1\) and \(x_i=0+i\Delta x\) for \(i=1,2,...,10\). Then
Applied Differential Equations![[Picture]](lecture27.online-6cde5aed9a59077ada83b62ea1df445a.svg)