5.16 Euler's methodIB Maths: Applications and Interpretation HL: Revision notes

Many differential equations have no neat formula for the solution. Euler's method builds a numerical approximation instead. We are given a first-order equation $\frac{dy}{dx}=f(x,y)$ and a starting point $(x_0,y_0)$, called the initial condition. At any point the equation gives the gradient of the solution curve, so we follow the tangent for a short horizontal distance $h$, called the step length, then recalculate the gradient and repeat. The result is a list of points $(x_0,y_0),(x_1,y_1),(x_2,y_2),\ldots$ that approximate the true curve, joined by short straight segments.

For $\frac{dy}{dx}=f(x,y)$ with $y(x_0)=y_0$ and step length $h$: $x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\,f(x_n,y_n).$ Example: $\frac{dy}{dx}=x+y$, $y(0)=1$, $h=0.1$.

• $y_1=1+0.1(0+1)=1.1$ at $x=0.1$.
• $y_2=1.1+0.1(0.1+1.1)=1.22$ at $x=0.2$.
• $y_3=1.22+0.1(0.2+1.22)=1.362$ at $x=0.3$. So $y(0.3)\approx1.36$. The true value is $1.40$ (3 s.f.), so the estimate is close but not exact.

Repeated steps are slow by hand, so in examinations the values are generated using permissible technology. In a spreadsheet, put $x_0$ and $y_0$ in the first row. In the next row type the recurrences, for example =A2+0.1 for $x$ and =B2+0.1*(A2+B2) for $y$, then fill down. Most GDCs can do the same with a recursive sequence or table mode. Keep full precision in the calculator and round only the final answer to 3 significant figures, unless the question says otherwise. A spreadsheet is also the quickest way to answer 'when does $y$ first exceed...?' questions: read down the column.

Each step replaces a curve by a tangent, so a small error is made, and errors build up over many steps. Using a smaller step length $h$ gives a more accurate estimate but needs more steps. Halving $h$ roughly halves the final error. If the true solution is concave up (gradient increasing), the tangent lies below the curve and Euler's method underestimates. If it is concave down, Euler's method overestimates. In the example above the gradient $x+y$ increases, so $1.362<1.40$.

Two quantities that depend on each other, both changing with time, give a coupled system: $\frac{dx}{dt}=f_1(x,y,t)$ and $\frac{dy}{dt}=f_2(x,y,t)$. Euler's method now steps $t$, $x$ and $y$ together: $t_{n+1}=t_n+h,\quad x_{n+1}=x_n+h\,f_1(x_n,y_n,t_n),\quad y_{n+1}=y_n+h\,f_2(x_n,y_n,t_n).$ Both new values use the OLD $x_n$ and $y_n$. Example: $\frac{dx}{dt}=-y$, $\frac{dy}{dt}=x$, $x_0=1$, $y_0=0$, $h=0.1$. Then $x_1=1+0.1(0)=1$ and $y_1=0+0.1(1)=0.1$. Next $x_2=1+0.1(-0.1)=0.99$ and $y_2=0.1+0.1(1)=0.2$.

A common coupled model has prey $x$ and predators $y$: $\frac{dx}{dt}=ax-bxy,\qquad \frac{dy}{dt}=cxy-dy.$ The term $ax$ is prey growth with plenty of food, $-bxy$ is prey lost to predators (it needs both to meet), $cxy$ is predator growth from eating prey, and $-dy$ is predator deaths. The populations are constant when both rates are zero: $y=\frac{a}{b}$ and $x=\frac{d}{c}$ (the non-zero equilibrium). Example: $\frac{dx}{dt}=0.6x-0.03xy$, $\frac{dy}{dt}=0.02xy-0.5y$ with $x=50$, $y=10$: the rates are $15$ and $5$, so with $h=0.2$ we get $x_1=53$, $y_1=11$. Euler's method on a spreadsheet then shows both populations rising and falling in cycles, with the predators peaking after the prey.