Euler's Method

A tangent line approximates a solution curve well only near its point of tangency. Euler's method takes several short tangent-line steps instead, and finds a new slope from the differential equation at the start of each one.

The method

Each step uses the slope at its starting point, so the steps usually drift away from the solution curve as they go. Smaller steps stay closer to it, at the cost of more steps. A negative

\Delta x

steps to the left. On a free-response answer, show each step with the slope it uses.

Example 1

Let

y = f(x)

be the solution of

\frac{dy}{dx} = (2x - y)^2 + 2

with

f(0) = -1

(a) Use Euler's method with steps of

\Delta x = 0.5

, starting at

x = 0

, to approximate

f(1)

(b) Use the sign of

\frac{d^2y}{dx^2}

to decide whether the approximation is too high or too low.

f(x) = 2x - \frac{1}{x + 1}

, and compare

f(1)

with the approximation.

(a) Two steps of

0.5

reach

x = 1

$x$	$y$	Slope $(2x - y)^2 + 2$
$0$	$-1$	$1 + 2 = 3$
$0.5$	$0.5$	$0.25 + 2 = 2.25$
$1$	$1.625$

The steps are

-1 + 3(0.5) = 0.5

and

0.5 + 2.25(0.5) = 1.625

f(1) \approx 1.625.

(b) Differentiate with respect to

x

, using the chain rule, and substitute for

\frac{dy}{dx}

\begin{aligned} \frac{d^2y}{dx^2} &= 2(2x - y)\left(2 - \frac{dy}{dx}\right) \\[4pt] &= 2(2x - y)\left(-(2x - y)^2\right) \\[4pt] &= -2(2x - y)^3. \end{aligned}

So every solution curve is concave down where

2x - y > 0

. That holds at both starting points,

(0, -1)

and

(0.5, 0.5)

, and along the solution curves through them up to

x = 1

. Each step follows a tangent line of a concave-down curve, so it runs above that curve, and

1.625

is too high.

(c) For

y = 2x - \frac{1}{x + 1}

, the derivative is

2 + \frac{1}{(x + 1)^2}

. Since

2x - y = \frac{1}{x + 1}

, the right side is

\frac{1}{(x + 1)^2} + 2

, the same. Also

f(0) = 0 - 1 = -1

, so it's

f

, on

x > -1

Then

f(1) = 2 - \frac{1}{2} = 1.5

, below the approximation

1.625

, as (b) said. Four steps of

0.25

give about

1.550

, which is closer.

The solution y = 2x − 1/(x + 1), the two-step Euler approximation (amber), and the four-step one (dashed blue). The steps run above the concave-down curve, and the smaller steps stay closer to it.

Euler's method in context

Example 2

In a laboratory reaction, an enzyme breaks down a substrate. The concentration

S

of the substrate, in millimoles per liter, satisfies

\frac{dS}{dt} = -\frac{6S}{2 + S},

where

t

is measured in hours. A sample taken at

t = 1

shows

S(1) = 1

(a) Use Euler's method with

\Delta t = -0.5

, starting at

t = 1

, to estimate the starting concentration

S(0)

(b) Use the sign of

\frac{d^2S}{dt^2}

to decide whether the estimate is too high or too low.

(a) Each step changes

t

-0.5

, so it changes

S

by the slope times

-0.5

$t$	$S$	Slope $-\frac{6S}{2 + S}$
$1$	$1$	$-\frac{6}{3} = -2$
$0.5$	$2$	$-\frac{12}{4} = -3$
$0$	$3.5$

The steps are

1 + (-2)(-0.5) = 2

and

2 + (-3)(-0.5) = 3.5

. So

S(0) \approx 3.5

millimoles per liter.

(b) By the quotient rule and the chain rule,

\begin{aligned} \frac{d^2S}{dt^2} &= -\frac{6(2 + S) - 6S}{(2 + S)^2} \cdot \frac{dS}{dt} \\[4pt] &= -\frac{12}{(2 + S)^2} \cdot \left(-\frac{6S}{2 + S}\right) \\[4pt] &= \frac{72S}{(2 + S)^3}. \end{aligned}

This is positive whenever

S > 0

, so every solution with positive concentration is concave up. A tangent line lies below a concave-up curve on both sides of its point of tangency. So a step to the left ends below the curve too, and the estimate

3.5

is too low.

The true value is

S(0) \approx 4.15

. Four steps of

-0.25

give about

3.81

The concentration S(t) through (1, 1) and the two backward Euler steps (amber). Both steps end below the concave-up curve.