Euler's methods for differential equationsAQA A-Level Further Maths: Revision notes

A first order differential equation $\frac{dy}{dx}=f(x,y)$ with a starting point $(x_0,y_0)$ can be solved step by step when no exact solution is available. Euler's method follows the tangent at the current point for a step of width $h$: $y_{r+1}=y_r+hf(x_r,y_r),\qquad x_{r+1}=x_r+h.$ The gradient is worked out at the start of each step and then treated as constant across it.

$\frac{dy}{dx}=x+y$, $y(0)=1$, $h=0.1$. Step 1: $y_1=1+0.1(0+1)=1.1$ at $x_1=0.1$. Step 2: $y_2=1.1+0.1(0.1+1.1)=1.22$ at $x_2=0.2$. So $y(0.2)\approx1.22$. The exact solution $y=2e^{x}-x-1$ gives $y(0.2)=1.2428$, so the estimate is a little low. Present results in a table of $x_r$, $y_r$ and $f(x_r,y_r)$.

The improved Euler method uses the gradient at the midpoint of two steps, which is more accurate: $y_{r+1}=y_{r-1}+2hf(x_r,y_r),\qquad x_{r+1}=x_r+h.$ It needs two starting values, $y_0$ and $y_1$, so the first step uses ordinary Euler. Example. For $\frac{dy}{dx}=x+y$ with $y_0=1$, $y_1=1.1$: $y_2=y_0+2hf(x_1,y_1)=1+0.2(0.1+1.1)=1.24$, which is closer to the exact $1.2428$ than the ordinary Euler value $1.22$.

Euler's method follows tangent lines, so the error grows with each step. Reducing $h$ makes it more accurate, but more steps are needed. If the exact curve is convex ($y''>0$) the tangents lie below it and Euler's method underestimates; if concave it overestimates. Because the improved Euler method uses the gradient at the middle of the interval it is usually much more accurate for the same $h$. Measure accuracy with the percentage error $\frac{|\text{estimate}-\text{exact}|}{\text{exact}}\times100$.