Euler's Method

Most differential equations can’t be solved with a neat formula. Euler’s method (say “OY-ler”) gets a numerical answer anyway: start at a known point, follow the tangent line a short distance, recompute the slope from the differential equation, and repeat. It is linearization used over and over, and it’s the numerical cousin of following a slope field. On the AP BC exam it usually appears as a short no-calculator part of a differential equations question.

Key ideas

One step

Suppose $\dfrac{dy}{dx} = f(x, y)$ and you know the point $(x_0, y_0)$ is on the solution. The slope there is $f(x_0, y_0)$. Move right by a step size $\Delta x$ along the tangent line:

$$y_1 = y_0 + f(x_0, y_0)\,\Delta x, \qquad x_1 = x_0 + \Delta x$$

In words: new $y$ = old $y$ + slope × step.

Repeat

At the new point $(x_1, y_1)$, compute a new slope from the differential equation and step again. A table keeps the work organized:

$x$	$y$	$\dfrac{dy}{dx}$ at this point	$\Delta y = \dfrac{dy}{dx}\cdot\Delta x$
$x_0$	$y_0$	$f(x_0, y_0)$	$f(x_0, y_0)\,\Delta x$
$x_1$	$y_1$	$f(x_1, y_1)$	$f(x_1, y_1)\,\Delta x$
$\vdots$	$\vdots$

To reach a target $x$ from $x_0$ with step $\Delta x$, you need $\dfrac{\text{target} - x_0}{\Delta x}$ steps.

Step size

Each step follows a tangent line, which drifts away from the true curve. Smaller steps stay closer to the curve and usually give better estimates, at the cost of more arithmetic.

Too high or too low?

The answer depends on the concavity of the solution:

• If the solution is concave up ($\dfrac{d^2y}{dx^2} \gt 0$), tangent lines lie below the curve, so Euler’s method gives an underestimate.
• If the solution is concave down ($\dfrac{d^2y}{dx^2} \lt 0$), tangent lines lie above the curve, so Euler’s method gives an overestimate.

You can find $\dfrac{d^2y}{dx^2}$ without solving the equation: differentiate $\dfrac{dy}{dx}$ implicitly, then substitute the differential equation in for every $\dfrac{dy}{dx}$. On the AP exam, justify with a sentence like ”$\dfrac{d^2y}{dx^2} \gt 0$ on this interval, so the solution curve is concave up and lies above its tangent lines; the Euler approximation is an underestimate.”

Worked examples

Example 1: Two steps

Let $y = f(x)$ be the solution of $\dfrac{dy}{dx} = x + y$ with $f(0) = 1$. Use Euler’s method with two steps of equal size to approximate $f(1)$.

Solution. Two steps from $x = 0$ to $x = 1$ means $\Delta x = 0.5$.

$x$	$y$	$\dfrac{dy}{dx} = x + y$	$\Delta y = \dfrac{dy}{dx}(0.5)$
$0$	$1$	$1$	$0.5$
$0.5$	$1.5$	$2$	$1$
$1$	$2.5$

So $f(1) \approx 2.5$.

Example 2: A smaller step, and under or over?

For the same equation, (a) approximate $f(1)$ with four steps, (b) decide whether these approximations are under- or overestimates. (c) The exact solution is $f(x) = 2e^x - x - 1$. Compare.

Solution.

(a) Now $\Delta x = 0.25$:

$x$	$y$	$\dfrac{dy}{dx} = x + y$	$\Delta y = \dfrac{dy}{dx}(0.25)$
$0$	$1$	$1$	$0.25$
$0.25$	$1.25$	$1.5$	$0.375$
$0.5$	$1.625$	$2.125$	$0.53125$
$0.75$	$2.15625$	$2.90625$	$0.7265625$
$1$	$2.8828125$

So $f(1) \approx 2.883$.

(b) Differentiate the differential equation:

$$\frac{d^2y}{dx^2} = 1 + \frac{dy}{dx} = 1 + x + y$$

Starting from $(0, 1)$, $x$ and $y$ both stay positive (the slope $x + y$ is positive, so $y$ keeps increasing). So $\dfrac{d^2y}{dx^2} = 1 + x + y \gt 0$ for $0 \le x \le 1$: the solution is concave up, its tangent lines lie below it, and both Euler approximations are underestimates.

(c) Exactly, $f(1) = 2e - 2 \approx 3.437$. The two-step estimate is off by about $0.937$; the four-step estimate is off by about $0.554$. Halving the step size roughly halved the error.

Euler’s method for $\frac{dy}{dx} = x + y$, $y(0) = 1$: the polygons stay below the concave-up solution, and the smaller step gets closer.

Example 3: A cooling problem that overestimates

A cold drink at $4\,^\circ\text{C}$ is set out in a $22\,^\circ\text{C}$ room. Its temperature $T$ (in $^\circ\text{C}$) after $t$ minutes satisfies $\dfrac{dT}{dt} = -0.5(T - 22)$. (a) Use Euler’s method with $\Delta t = 1$ to estimate $T(2)$. (b) Is your estimate too high or too low?

Solution.

(a)

$t$	$T$	$\dfrac{dT}{dt} = -0.5(T - 22)$	$\Delta T = \dfrac{dT}{dt}(1)$
$0$	$4$	$9$	$9$
$1$	$13$	$4.5$	$4.5$
$2$	$17.5$

So $T(2) \approx 17.5\,^\circ\text{C}$.

(b) Differentiate:

$$\frac{d^2T}{dt^2} = -0.5\,\frac{dT}{dt} = -0.5\big(-0.5(T - 22)\big) = 0.25(T - 22)$$

The drink stays colder than the room ($T \lt 22$), so $\dfrac{d^2T}{dt^2} \lt 0$. The solution is concave down, the tangent lines lie above it, and $17.5$ is an overestimate.

(The exact solution is $T = 22 - 18e^{-0.5t}$, so $T(2) \approx 15.378\,^\circ\text{C}$, which confirms it.)

Common mistakes

Using the same slope for every step. The whole point is to recompute $\dfrac{dy}{dx}$ at each new point, using the new $x$ and the new $y$.

Forgetting to multiply by the step size. The change is $\Delta y = \dfrac{dy}{dx}\cdot\Delta x$, not just $\dfrac{dy}{dx}$. With $\Delta x = 0.5$, a slope of $2$ only raises $y$ by $1$.

Counting steps wrong. From $x = 0$ to $x = 1$ with $\Delta x = 0.25$ is four steps, and the slope at the last point ($x = 1$) is never used.

Mixing up concave up and under/over. Concave up means the curve bends above its tangent lines, so the tangent-line steps fall short: an underestimate. Picture $y = x^2$ and its tangent lines if you’re unsure.

Forgetting to substitute when finding the second derivative. $\dfrac{d}{dx}(x + y) = 1 + \dfrac{dy}{dx}$, and then you must replace $\dfrac{dy}{dx}$ with $x + y$. An answer that still contains $\dfrac{dy}{dx}$ can’t be evaluated at a point.

Practice

1. (Warm-up) Let $y = f(x)$ satisfy $\dfrac{dy}{dx} = 2x$ with $f(0) = 1$. Use Euler’s method with $\Delta x = 0.5$ to approximate $f(1)$.

Solution

$x$	$y$	$\dfrac{dy}{dx} = 2x$	$\Delta y$
$0$	$1$	$0$	$0$
$0.5$	$1$	$1$	$0.5$
$1$	$1.5$

$f(1) \approx 1.5$. (The exact solution $y = x^2 + 1$ gives $2$; since $y'' = 2 \gt 0$, the estimate is low, as expected.)

2. (Warm-up) Given $\dfrac{dy}{dx} = y^2 - x$ and $y(1) = 2$, use one step of Euler’s method with $\Delta x = 0.1$ to approximate $y(1.1)$.

Solution

The slope at $(1, 2)$ is $2^2 - 1 = 3$, so

$$y(1.1) \approx 2 + 3(0.1) = 2.3$$

3. (Warm-up) Given $\dfrac{dy}{dx} = x - y$ and $y(0) = 2$, use Euler’s method with two equal steps to approximate $y(1)$.

Solution

$\Delta x = 0.5$.

$x$	$y$	$\dfrac{dy}{dx} = x - y$	$\Delta y$
$0$	$2$	$-2$	$-1$
$0.5$	$1$	$-0.5$	$-0.25$
$1$	$0.75$

$y(1) \approx 0.75$.

4. (Core) Let $y = f(x)$ be the solution of $\dfrac{dy}{dx} = xy$ with $f(0) = 1$.

• (a) Use Euler’s method with $\Delta x = 0.25$ to approximate $f(1)$. Give your answer to 3 decimal places.
• (b) Find $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$, and decide whether your answer to (a) is an under- or overestimate.

Solution

(a)

$x$	$y$	$\dfrac{dy}{dx} = xy$	$\Delta y = \dfrac{dy}{dx}(0.25)$
$0$	$1$	$0$	$0$
$0.25$	$1$	$0.25$	$0.0625$
$0.5$	$1.0625$	$0.53125$	$0.1328125$
$0.75$	$1.1953125$	$0.896484375$	$0.22412109375$
$1$	$1.41943359375$

$f(1) \approx 1.419$.

(b) By the product rule, $\dfrac{d^2y}{dx^2} = y + x\dfrac{dy}{dx} = y + x(xy) = y\left(1 + x^2\right)$.

The slope $xy$ is never negative for $x \ge 0$ and $y \gt 0$, so $y$ never drops below $1$. Then $\dfrac{d^2y}{dx^2} = y(1 + x^2) \gt 0$: the solution is concave up, so $1.419$ is an underestimate. (The exact value is $e^{1/2} \approx 1.649$.)

5. (Core) Given $\dfrac{dy}{dx} = y - x$ and $y(0) = 2$:

• (a) Use Euler’s method with $\Delta x = 0.2$ to approximate $y(0.4)$.
• (b) Show that $\dfrac{d^2y}{dx^2} = y - x - 1$. Given that the solution satisfies $y \gt x + 1$ for $0 \le x \le 0.4$, is your estimate too high or too low?

Solution

(a)

$x$	$y$	$\dfrac{dy}{dx} = y - x$	$\Delta y$
$0$	$2$	$2$	$0.4$
$0.2$	$2.4$	$2.2$	$0.44$
$0.4$	$2.84$

$y(0.4) \approx 2.84$.

(b) $\dfrac{d^2y}{dx^2} = \dfrac{dy}{dx} - 1 = (y - x) - 1$. If $y \gt x + 1$, then $y - x - 1 \gt 0$, so the solution is concave up and $2.84$ is too low. (The exact solution $y = x + 1 + e^x$ gives $y(0.4) \approx 2.892$.)

6. (Core) Water drains from a tank. The depth $h$ in metres after $t$ minutes satisfies $\dfrac{dh}{dt} = -0.2\sqrt{h}$, with $h(0) = 9$.

• (a) Use Euler’s method with $\Delta t = 1$ to approximate $h(2)$. Give your answer to 3 decimal places.
• (b) Is the approximation an under- or overestimate? Justify.

Solution

(a) At $t = 0$: slope $= -0.2\sqrt{9} = -0.6$, so $h(1) \approx 9 - 0.6 = 8.4$.

At $t = 1$: slope $= -0.2\sqrt{8.4} \approx -0.5797$, so $h(2) \approx 8.4 - 0.5797 \approx 7.820$ m.

(b) By the chain rule,

$$\frac{d^2h}{dt^2} = -0.2 \cdot \frac{1}{2\sqrt{h}} \cdot \frac{dh}{dt} = -\frac{0.1}{\sqrt{h}}\left(-0.2\sqrt{h}\right) = 0.02 \gt 0$$

The solution is concave up, so the estimate is an underestimate. (The exact solution is $h = (3 - 0.1t)^2$, giving $h(2) = 7.84$.)

7. (Core) A quantity satisfies $\dfrac{dy}{dt} = ky$ with $y(0) = 2$. One step of Euler’s method with $\Delta t = 0.5$ gives $y(0.5) \approx 3$. Find $k$.

Solution

The slope at $(0, 2)$ is $2k$, so one step gives

$$2 + 2k(0.5) = 3 \quad\Rightarrow\quad 2 + k = 3 \quad\Rightarrow\quad k = 1$$

8. (Challenge) Let $y = f(x)$ satisfy $\dfrac{dy}{dx} = (y - 1)x^2$ with $f(0) = 3$.

• (a) Use Euler’s method with two equal steps to approximate $f(1)$.
• (b) Find $\dfrac{d^2y}{dx^2}$ and explain why the approximation is an underestimate.

Solution

(a) $\Delta x = 0.5$. At $(0, 3)$: slope $= (2)(0) = 0$, so $y(0.5) \approx 3$. At $(0.5, 3)$: slope $= (2)(0.25) = 0.5$, so $f(1) \approx 3 + 0.5(0.5) = 3.25$.

(b) By the product rule,

$$\frac{d^2y}{dx^2} = \frac{dy}{dx}\,x^2 + (y - 1)(2x) = (y - 1)x^4 + 2x(y - 1) = (y - 1)\left(x^4 + 2x\right)$$

Starting at $y = 3$, the slope $(y - 1)x^2$ is never negative while $y \gt 1$, so $y$ stays at least $3$. For $0 \lt x \le 1$, both $y - 1 \gt 0$ and $x^4 + 2x \gt 0$, so $\dfrac{d^2y}{dx^2} \gt 0$. The solution is concave up, so $3.25$ is an underestimate. (The exact value is $1 + 2e^{1/3} \approx 3.791$.)

9. (Challenge) For $\dfrac{dy}{dx} = 3 - y$ with $y(0) = 1$, show that Euler’s method with $n$ equal steps from $x = 0$ to $x = 1$ gives $y(1) \approx 3 - 2\left(1 - \tfrac{1}{n}\right)^n$. What happens as $n \to \infty$?

Solution

Let $h = \tfrac{1}{n}$. One step gives $y_{k+1} = y_k + (3 - y_k)h$. Subtract both sides from $3$:

$$3 - y_{k+1} = (3 - y_k) - (3 - y_k)h = (3 - y_k)(1 - h)$$

So the gap $3 - y$ is multiplied by $(1 - h)$ at each step. It starts at $3 - 1 = 2$, so after $n$ steps it is $2(1 - h)^n$, and

$$y(1) \approx 3 - 2\left(1 - \tfrac{1}{n}\right)^n$$

As $n \to \infty$, $\left(1 - \tfrac{1}{n}\right)^n \to e^{-1}$, so the estimates approach $3 - \dfrac{2}{e} \approx 2.264$. That is the exact value: the solution is $y = 3 - 2e^{-x}$. Smaller and smaller steps converge to the true answer.