6.1.3 Practice set: first-order equations

Separable equations and Euler's method

Differential equations and separable equations (lesson 6.1.1)

0 of 9 correct

1.
For $y' = y - t^2$ , what slope does the slope field show at the point $(1, 3)$ ?
2.
Solve $y' = 0.25y$ with $y(0) = 8$ , and give $y(4)$ (type e for $e$ )
3.
How long does $y' = 0.07y$ take to double $y$ ? (type ln for the natural log)
4.
Solve $y' = 0.5(10 - y)$ with $y(0) = 4$ . Write $y(t)$
5.
Solve $y' = 3t^2y$ with $y(0) = 2$ . Write $y(t)$
6.
Solve $y' = -y^2$ with $y(0) = 1$ , and give $y(3)$
7.
Take two Euler steps of $h = 0.25$ for $y' = -2y$ from $y(0) = 1$ . What is $y$ at $t = 0.5$ ?
8.
One Euler step of $h = 5$ for $y' = 0.5(10 - y)$ from $y(0) = 4$ lands where?
9.
Separating $y' = y(y - 3)$ divides by $y(y - 3)$ . Which positive constant solution does that lose?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Linear equations and `solve_ivp`

Linear equations and integrating factors (lesson 6.1.2)

0 of 9 correct

1.
Write $3y' - 6y = 12$ in standard form. What is $p$ ?
2.
Find the integrating factor of $y' - 0.5y = t$
3.
Find the integrating factor of $y' + \frac{2y}{t} = 1$ for $t > 0$
4.
Solve $y' + 3y = 6$ with $y(0) = 5$ . Write $y(t)$
5.
Every solution of $y' + y = 2t$ approaches one straight line. Write it
6.
Solve $y' + \frac{2y}{t} = 1$ with $y(1) = 1$ , and give $y(2)$
7.
Solve $y' - y = 1$ with $y(0) = 0$ , and give $y(1)$ (type e for $e$ )
8.
With the defaults rtol = $10^{-3}$ and atol = $10^{-6}$ , solve_ivp keeps each step's error below about atol + rtol × |y|. Where $|y| = 0.002$ , how many millionths ( $10^{-6}$ ) is that?
9.
The solution $50e^{-2t}$ falls below the default atol of $10^{-6}$ at what time? (type ln for the natural log)

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Coding exercise

Exercise

Euler's method, and a step that halves itself

Write two functions for an equation $y' = f(t, y)$ , where f(t, y) returns the slope.

euler(f, t0, y0, t_end, n) takes n Euler steps from $(t_0, y_0)$ to t_end, each of length $h = \frac{t_{\text{end}} - t_0}{n}$ . It returns two NumPy arrays of length $n + 1$ : the times $t_0 + kh$ and the values $y_k$ . If t_end is before t0, $h$ is negative and the steps run backwards. Raise ValueError unless n is a positive whole number.
euler_converge(f, t0, y0, t_end, tol, n0=10) starts with n0 steps and doubles the number until two answers in a row at t_end differ by less than tol. Return the last answer and its number of steps, as a pair.

Euler's error is about proportional to $h$ (lesson 6.1.1), so the gap between two answers in a row is about the error of the second. For $y' = 0.5y$ from $y(0) = 1$ to $t = 2$ with tol=1e-3, the second function stops at 2,560 steps, about 0.00053 below $e$ .

See the tests

import math

import numpy as np


def growth(t, y):
    return 0.5 * y


def test_four_steps():
    """Four steps of 0.5 for y' = 0.5y from 1 give 1.25^4 = 2.441406"""
    ts, ys = euler(growth, 0, 1, 2, 4)
    expect_close(ys[-1], 1.25**4, what="y after 4 steps")


def test_arrays():
    """The times run from t0 to t_end in n + 1 values, with y0 first"""
    ts, ys = euler(growth, 0, 1, 2, 4)
    expect_equal(len(ts), 5, what="len(ts)")
    expect_equal(len(ys), 5, what="len(ys)")
    expect_close(ts, [0, 0.5, 1, 1.5, 2], what="ts")
    expect_close(ys[0], 1.0, what="ys[0]")


def test_slope_depends_on_t():
    """Two steps of 0.1 for y' = t + y from y(0) = 1 give 1.1, then 1.22"""
    ts, ys = euler(lambda t, y: t + y, 0, 1, 0.2, 2)
    expect_close(ys, [1, 1.1, 1.22], what="ys")


def test_backwards():
    """2,000 steps back from y(2) = e to t = 0 for y' = 0.5y land near 1"""
    ts, ys = euler(growth, 2, math.e, 0, 2000)
    expect_close(ts[-1], 0.0, what="last time")
    expect_close(ys[-1], 1.0, what="y at t = 0", rel=1e-3)


def test_bad_n():
    """n = 0 and n = 2.5 raise ValueError"""
    for n in (0, 2.5):
        try:
            euler(growth, 0, 1, 2, n)
        except ValueError:
            continue
        raise AssertionError(f"expected ValueError for n = {n}")


def test_converge():
    """For y' = 0.5y to t = 2 with tol 0.001, the steps double to 2,560 and land 0.00053 below e"""
    value, n = euler_converge(growth, 0, 1, 2, 1e-3)
    expect_equal(n, 2560, what="number of steps")
    expect_close(value, math.e, what="answer", rel=3e-4)


def test_converge_settling():
    """For y' = 0.8(2 - y) from 4 to t = 5 with tol 10^-5, the answer is within 10^-5 of 2 + 2e^-4"""
    value, n = euler_converge(lambda t, y: 0.8 * (2 - y), 0, 4, 5, 1e-5)
    expect_close(value, 2 + 2 * math.exp(-4), what="answer", rel=0, abs_tol=1e-5)

Separable equations and Euler's method

Linear equations and solve_ivp

Coding exercise

Free further reading

Linear equations and `solve_ivp`