Slope fields and Euler’s method

A differential equation gives the slope of the solution through every point. You can check a proposed solution by substituting it, draw the slopes as a field of short segments and read the shape of every solution at once, or step along them with Euler’s method and say whether the estimate runs high or low.

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In this topic

Taught: Before the final, flavours A and B
Questions: 14
Total marks: 36
Suggested time: 65 min
Difficulty: Introductory to Capstone
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Key results

Checking a solution: $y = \varphi(t) \text{ solves } \frac{dy}{dt} = f(t, y) \iff \varphi'(t) = f\bigl(t, \varphi(t)\bigr)$
For every $t$ in the interval considered; one matching point is not enough.
Linear differential equations: $\alpha(t)\,\frac{dy}{dt} + \beta(t)\,y + \gamma(t) = 0$
The coefficients may depend on $t$ but not on $y$ ; an equation that cannot be written this way is nonlinear.
Euler’s method: $y_{n+1} = y_{n} + \Delta t\, f(t_{n}, y_{n}), \qquad t_{n+1} = t_{n} + \Delta t$
For $\dfrac{dy}{dt} = f(t, y)$ with $y(t_{0}) = y_{0}$ ; then $y_{n}$ estimates $y(t_{n})$ .
Concavity from the equation: $\frac{dy}{dt} = f(y) \implies \frac{d^{2}y}{dt^{2}} = f'(y)\,f(y)$
When $f$ is differentiable.
Euler’s method runs low or high: $y'' > 0 \implies y_{n} < y(t_{n}), \qquad y'' < 0 \implies y_{n} > y(t_{n})$
For $n \ge 1$ and $\Delta t > 0$ , when every solution through the region the steps cross is concave up (for the first) or concave down (for the second).

Read the theoryUBC Math 100 §11.1 Introducing a new kind of equationUBC Math 100 §11.2 Verifying that a function is a solutionUBC Math 100 §12.1 Linear and nonlinear differential equationsUBC Math 100 §12.2 Slope fieldsUBC Math 100 §12.3 Euler’s method

Worked example

Standard3 marks · 5 min

Early in the season an insect population grows at a per-capita rate that falls as the season goes on. With $P$ in thousands of insects and $t$ in weeks,

\frac{dP}{dt} = (0.6 - 0.2t)\,P, \qquad P(0) = 100.

Use Euler’s method with two steps of $\Delta t = 1$ to estimate $P(2)$ .

Strategy

Step along the tangent twice. Each step uses the slope at the point it starts from, which needs both the current time and the current population.

Working

$f(0, 100) = (0.6)(100) = 60, \qquad P_{1} = 100 + 1\cdot 60 = 160$
First step, from t = 0.
$f(1, 160) = (0.6 - 0.2)(160) = 64, \qquad P_{2} = 160 + 1\cdot 64 = 224$
Second step, from t = 1: both t and P have moved on.
$P(t) = 100e^{0.6t - 0.1t^{2}} \implies P(2) = 100e^{0.8} \approx 222.6$
For comparison: differentiating shows this formula solves the problem, and the two-step estimate is close to it.

Answer

P(2) \approx P_{2} = 224 \text{ thousand}

Check

Over two weeks the per-capita rate averages 0.4, so P grows by a factor of about e^0.8 ≈ 2.23, to roughly 223 thousand, close to 224.

Where this goes wrong

Leaving t = 0 in the slope for the second step, which gives 256. By t = 1 the per-capita rate has fallen to 0.4, and the second step must use it.

On the paper

Lay the steps out as a table of $t$ , $y$ , the slope and the next $y$ , state the step size, and round only the final answer.

Leans on

Linear approximation, from Linear approximation

Your turn

Introductory2 marks · 3 min

A loaf of bread comes out of the oven at 95 °C into a kitchen at 25 °C. Its temperature $T(t)$ , in °C after $t$ minutes, satisfies $\dfrac{dT}{dt} = 0.05\,(25 - T)$ . Use one step of Euler’s method with $\Delta t = 2$ to estimate the temperature after 2 minutes.

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Practice questions

5 of the 14, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
In each equation $y$ is an unknown function of $t$ . Which one of these differential equations is nonlinear?
Strategy
Try to write each equation as α dy/dt + βy + γ = 0. The coefficients may depend on t but not on y; an equation that cannot be put in this form is nonlinear.
Working
$\frac{dy}{dt} + 2y - 6 = 0: \quad \alpha = 1,\ \beta = 2,\ \gamma = -6$
Constant coefficients: linear.
$\frac{dy}{dt} - t^{2}y - 1 = 0: \quad \alpha = 1,\ \beta = -t^{2},\ \gamma = -1$
A coefficient that depends on t is allowed: linear.
$2\frac{dy}{dt} + 3y - e^{t} = 0: \quad \alpha = 2,\ \beta = 3,\ \gamma = -e^{t}$
The e^t contains no y: linear.
$\frac{dy}{dt} = 6 - 2y^{2}$
The y² term is not a multiple of y by anything free of y, so this one is nonlinear.
Answer
$\dfrac{dy}{dt} = 6 - 2y^{2}$
The answer is dy by dt equals 6 minus 2 y squared.
Check
6 − 2y² is zero at y = √3 and at y = −√3 but not at y = 0, which no linear expression a + by can do.
Where this goes wrong
Calling an equation nonlinear because t appears squared or in an exponential. Linearity is about y alone: t may appear in any form, but y and dy/dt must appear only to the first power.
Leans on
What a differential equation is, from Exponential growth and decay
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The table gives the slope of the segment drawn at each point $(t, y)$ of a slope field. Which differential equation was the field drawn for?
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Introductory2 marks · 3 min
A drug is given by a steady intravenous drip and cleared by the body at a rate proportional to the amount present, so the amount $Q(t)$ , in mg after $t$ hours, satisfies $\dfrac{dQ}{dt} = 8 - 0.4Q$ . For exactly one value of the constant $b$ , the function $Q(t) = b + Ce^{-0.4t}$ is a solution for every constant $C$ . Find $b$ .
Strategy
To check a proposed solution, substitute it into both sides of the equation and compare. The exponential terms match for any C; the constant terms match for one value of b only.
Working
$\text{Left: } \frac{dQ}{dt} = -0.4Ce^{-0.4t}$
The constant b differentiates to 0, and the chain rule brings down −0.4.
$\text{Right: } 8 - 0.4\bigl(b + Ce^{-0.4t}\bigr) = 8 - 0.4b - 0.4Ce^{-0.4t}$
Substitute the whole of Q, constant term included.
$-0.4Ce^{-0.4t} = 8 - 0.4b - 0.4Ce^{-0.4t} \iff 8 - 0.4b = 0 \iff b = 20$
The exponential terms cancel and leave a condition on b alone. At 20 mg, clearance of 0.4 × 20 = 8 mg/h exactly balances the drip.
Answer
$b = 20 \text{ mg}$
The answer is b equals 20 milligrams.
Check
At Q = 20 mg the drip of 8 mg/h exactly balances clearance of 0.4 × 20 = 8 mg/h, so the amount holds steady there.
Where this goes wrong
Substituting only part of Q into the right-hand side, or losing the 8, so that the constant terms are never compared. Every term of the proposed solution goes into both sides.
Leans on
The derivative of e^kt, from The chain rule and logarithms
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Check that $y = \sqrt{t^{2} + C}$ solves $\dfrac{dy}{dt} = \dfrac{t}{y}$ for every positive constant $C$ . Then find the solution with $y(0) = 4$ , and give $y(3)$ .
Answer this question
Standard3 marks · 5 min
A mass on a frictionless spring is $x(t)$ centimetres from equilibrium after $t$ seconds, where
$\frac{d^{2}x}{dt^{2}} = -25x, \qquad x(0) = 10, \qquad x'(0) = 50.$
Find the positive $\omega$ for which $x(t) = A\cos(\omega t) + B\sin(\omega t)$ is a solution, use the initial conditions to find $A$ and $B$ , and give $x$ at $t = \tfrac{\pi}{20}$ .
Strategy
Substitute the proposed form into the equation to find which ω works, then let the two initial conditions fix the two constants, one each.
Working
$x' = -A\omega\sin(\omega t) + B\omega\cos(\omega t), \qquad x'' = -A\omega^{2}\cos(\omega t) - B\omega^{2}\sin(\omega t) = -\omega^{2}x$
Each differentiation brings out a factor ω by the chain rule.
$-\omega^{2}x = -25x \text{ for every } t \iff \omega^{2} = 25 \iff \omega = 5$
Take the positive root, as asked.
$x(0) = A = 10, \qquad x'(0) = 5B = 50 \implies B = 10$
At t = 0, sin 0 = 0 and cos 0 = 1, so each condition fixes one constant.
$x\left(\tfrac{\pi}{20}\right) = 10\cos\tfrac{\pi}{4} + 10\sin\tfrac{\pi}{4} = 10\sqrt{2} \approx 14.14$
Here 5t = π/4, where cos and sin both equal √2/2.
Answer
$x\left(\frac{\pi}{20}\right) = 10\sqrt{2} \approx 14.14 \text{ cm}$
The answer is 10 root 2 centimetres, about 14.14.
Check
The velocity −50 sin(5t) + 50 cos(5t) is 0 at t = π/20, so the mass is at its furthest point, and √(10² + 10²) ≈ 14.14 is the amplitude.
Where this goes wrong
Setting B = 50 straight from x′(0) = 50. Differentiating sin(5t) brings out a 5, so x′(0) = 5B and B = 10.
Leans on
The chain rule with sine and cosine, from The chain rule and logarithms
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A drug is given by a steady intravenous drip and cleared by the body at a rate proportional to the amount present, so the amount $Q(t)$ , in mg after $t$ hours, satisfies $\dfrac{dQ}{dt} = 8 - 0.4Q$ . For exactly one value of the constant $b$ , the function $Q(t) = b + Ce^{-0.4t}$ is a solution for every constant $C$ . Find $b$ .
Answer this question
Challenging3 marks · 5 min
Exactly one of these straight lines is itself a solution curve of $\dfrac{dy}{dt} = 2t - y$ . Which one?
Strategy
Substitute each line into both sides of the equation. A curve is a solution only if its slope agrees with the field at every point along it.
Working
$y = 2t - 2: \quad \frac{dy}{dt} = 2, \qquad 2t - y = 2t - (2t - 2) = 2 \quad\checkmark$
The two sides agree for every t.
$y = 2t: \quad \frac{dy}{dt} = 2, \qquad 2t - y = 0$
The field is flat along this line but the line climbs, so solutions cross it.
$y = t - 1: \quad \frac{dy}{dt} = 1, \qquad 2t - y = t + 1$
These agree only at t = 0; one matching point is not enough.
$y = 2t + 2: \quad \frac{dy}{dt} = 2, \qquad 2t - y = -2$
They disagree everywhere.
$y = 2t - 2 + Ce^{-t}$
Every other solution has this form (substitute to check), so all solutions approach the line y = 2t − 2.
Answer
$y = 2t - 2$
The answer is y equals 2 t minus 2.
Check
A line y = mt + c needs m = (2 − m)t − c for every t, which forces m = 2 and c = −2.
Where this goes wrong
Choosing y = 2t, the line where the slopes are zero. Zero slopes there mean solutions are momentarily flat as they cross it; the line itself rises, so it cannot be a solution.
Leans on
The slope of a straight line, from Functions, graphs and inverses
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The table gives the slope of the segment drawn at each point $(t, y)$ of a slope field. Which differential equation was the field drawn for?
Answer this question
Capstone5 marks · 10 min
A ball is dropped from rest and falls with air resistance proportional to its speed. Its downward velocity $v(t)$ , in m/s after $t$ seconds, satisfies
$\frac{dv}{dt} = 10 - 0.5v, \qquad v(0) = 0.$
Check that $v(t) = 20 + Ce^{-t/2}$ solves the equation for every constant $C$ , and choose $C$ to match the initial condition. Then estimate $v(2)$ with two Euler steps of $\Delta t = 1$ . Give the Euler estimate minus the true value $v(2)$ : positive if Euler overestimates, negative if it underestimates.
Strategy
Verify the family by substitution, let the initial condition choose C, take two Euler steps, and compare at t = 2. Concavity predicts the sign of the difference before any arithmetic.
Working
$v' = -\tfrac{1}{2}Ce^{-t/2}, \qquad 10 - 0.5\bigl(20 + Ce^{-t/2}\bigr) = -\tfrac{1}{2}Ce^{-t/2} \quad\checkmark$
Both sides agree for every C.
$v(0) = 20 + C = 0 \implies C = -20, \qquad v(t) = 20 - 20e^{-t/2}$
The ball starts from rest and approaches 20 m/s.
$v_{1} = 0 + 1\cdot(10 - 0) = 10, \qquad v_{2} = 10 + 1\cdot(10 - 5) = 15$
Each step uses the slope at the point it starts from.
$v(2) = 20 - 20e^{-1} = 20 - \frac{20}{e} \approx 12.64$
The true value from the formula.
$v_{2} - v(2) = 15 - 20 + \frac{20}{e} = \frac{20}{e} - 5 \approx 2.36$
Positive, as concavity predicts: v″ = −0.5v′ < 0 while the ball speeds up, so Euler’s tangent steps run above the curve.
Answer
$v_{2} - v(2) = \frac{20}{e} - 5 \approx 2.36 \text{ m/s}$
The answer is 20 over e minus 5, about 2.36 metres per second: Euler overestimates.
Check
Ten Euler steps of 0.2 give about 13.03 m/s, closer to the true 12.64 m/s but still above it, so the error is positive and shrinks with the step.
Where this goes wrong
Reporting 5 − 20/e, the true value minus the estimate. The sign carries the conclusion: the solution is concave down, so Euler’s estimate is too high and the difference asked for is positive.
Leans on
Equations of the form y′ = a − by, from Exponential growth and decay
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After an injection, a drug is removed by an enzyme that saturates, so its concentration $C(t)$ , in mg/L after $t$ hours, satisfies $\frac{dC}{dt} = -\frac{2C}{1 + C}, \qquad C(0) = 3.$
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Browse all 14 questions

Needs
Exponential growth and decay
The differential equations y′ = ky and y′ = a − by: half-lives, doubling times, Newton’s law of cooling and inflow and outflow.
11 questions
Leads to
Logistic growth and phase lines
Steady states and their stability from a state-space diagram (phase line), the logistic equation, and harvesting.
11 questions

Key results

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Working

Answer

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Related topics

Exponential growth and decay

Logistic growth and phase lines