What is the equation of the straight line with slope $m$ that passes through the point $(t,y) = (a,b)$?

Euler's method uses tangent lines as approximations to solution curves. The tangent line to a solution curve of $y'=f(t,y)$ at $(t_0,y_0)$ is $$ y=y_0+f(t_0,y_0)(t-t_0). $$

This approximates the curve when $t$ is close to $t_0$.

Now imagine a family of solution curves to the differential equation. We can calculate an approximate value for $y$ at some later time by taking lots of small steps in time. At each step we will use the tangent line to the solution through our current point. This is Euler's method.

Using $\Delta t$ as the step size for our algorithm, let

To find approximate $y$ values at these times, use the following equations in sequence:

and so on.

Example: Use Euler's method with $\Delta t=0.2,$ to find an approximate solution to $y(0.6)$ for the IVP \[ \dif{y}{t}=2t,\qquad y(0)=0. \] Compare your answer with the actual value.

Example: Use Euler's method with $\Delta t=0.2,$ to find an approximate solution to $y(0.6)$ for the IVP $\,\ds\dif{y}{t}=2t,$ $y(0)=0.\,$ Compare your answer with the actual value.

Example: Use Euler's method with $\Delta t=0.2,$ to find an approximate solution to $y(0.6)$ for the IVP $\,\ds\dif{y}{t}=2t,$ $y(0)=0.\,$ Compare your answer with the actual value.

The method is only accurate if you make $\Delta t$ small, which means a large number of steps is usually required. For this you might want to use Matlab. Define a $t$ vector and an initial $y$ value, then use the for command to iterate.

The method is only accurate if you make $\Delta t$ small, which means a large number of steps is usually required. For this you might want to use Matlab. Define a $t$ vector and an initial $y$ value, then use the for command to iterate.

In the example $\Delta t=0.05$ and 12 steps have been chosen to take $t$ to $0.6$.

t = (0:0.05:0.6); 
y(1) = 0;
for i = 1:12
  y(i+1) = y(i) + 2 * t(i) * 0.05;
end 
y(13)
plot(t, y,'-')

Using Matlab we find $y(0.6)=0.33$ which is much much better than the previous estimate value for $y(0.6)$.

Note that we start with $y(1)=0,$ whereas in the theoretical work we write $y(0)=0$. Matlab does not allow you to put an index of zero. Keep this in mind in case you ever get the Matlab error message:

???  Index into matrix is negative or zero.

Consider the IVP $ \ds\dif{y}{x}=\sin(xy),\; y(0)=1.\, $ Matlab code to estimate $y(2),$ using Euler's method with step size $\Delta x=0.01$ is:

									x = (0:0.01:2);
									y(1) = 1;
									for i = 1:200
									  y(i+1) = y(i) + sin(x(i) * y(i)) * 0.01;
									end 
									y(201)
									plot (x,y,'-')
								

Using Matlab we find $y(2)=1.8243.$

Use Euler's method to approximate the solution curve of the initial value problem

Use the slope field below as a guide to draw in the approximate solution curve. What goes wrong in this case?

Use Euler's method to approximate the solution curve of the initial value problem $ y' = y(1-y),\,$ $ y(-2) = 2. $ Use the slope field below as a guide to draw in the approximate solution curve. What goes wrong in this case?