6.2.3 Practice set: second-order equations

The characteristic equation

Constant coefficients and the characteristic equation (lesson 6.2.1)

0 of 8 correct

1.
$y'' - 7y' + 10y = 0$ has solutions $e^{rt}$ . Find the smaller $r$
2.
$3y'' - 2y' - y = 0$ has solutions $e^{rt}$ . Find the negative $r$
3.
$y'' + 10y' + 25y = 0$ has the solution $e^{-5t}$ . Write a second solution that is not a multiple of it
4.
Find the period of the waves that solve $y'' + 4y' + 13y = 0$ (type pi for $\pi$ )
5.
The waves of $y'' + 0.4y' + 4y = 0$ shrink like $e^{-0.2t}$ . After how long are they half as big? (type ln for the natural log)
6.
For which $c$ is $y'' + 6y' + cy = 0$ critically damped?
7.
$y'' + 2y' = 0$ has solutions $e^{rt}$ . Find the larger $r$
8.
Find the period of the waves that solve $y'' + 4y = 0$ (type pi for $\pi$ )

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.

General solutions and boundary conditions

General solutions and boundary conditions (lesson 6.2.2)

0 of 9 correct

1.
$y = c_1e^{2t} + c_2e^{-2t}$ solves $y'' - 4y = 0$ with $y(0) = 3$ and $y'(0) = 2$ . Find $c_1$
2.
Solve $y'' + 4y = 0$ with $y(0) = 1$ and $y'(0) = 4$ . Find $y(\frac{\pi}{4})$
3.
$y = (c_1 + c_2t)e^{-3t}$ solves $y'' + 6y' + 9y = 0$ with $y(0) = 2$ and $y'(0) = -1$ . Find $c_2$
4.
Solve $y'' + 4y = 0$ with $y(0) = 0$ and $y(\frac{\pi}{4}) = 3$ . Find $y(\frac{\pi}{12})$
5.
How many solutions has $y'' + 4y = 0$ with $y(0) = 1$ and $y(\pi) = 2$ ?
6.
Solve $x^2y'' + 3xy' - 3y = 0$ for $x \ge 1$ with $y(1) = 2$ and $y$ bounded as $x \to \infty$ . Find $y(2)$
7.
$y'' - y = 0$ from $y(0) = 0$ with slope $s$ ends at $y(1) = s\,\frac{e - e^{-1}}{2}$ . Find the slope that lands on $y(1) = 2$ (type e, or give 4 significant figures)
8.
$y = e^{-t}(c_1\cos 3t + c_2\sin 3t)$ solves $y'' + 2y' + 10y = 0$ with $y(0) = 3$ and $y'(0) = 0$ . Find $c_2$
9.
$x^2y'' - 12y = 0$ has solutions $x^p$ . Find the positive $p$

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

Solving any constant-coefficient equation

Write ivp_solution(a, b, c, y0, v0, t). It returns $y(t)$ for the solution of $ay'' + by' + cy = 0$ with $y(0) = y_0$ and $y'(0) = v_0$ , by the formulas of lessons 6.2.1 and 6.2.2. t may be a number or a NumPy array.

Decide the case from the discriminant $b^2 - 4ac$ , then write the general solution of that case.
Fix $c_1$ and $c_2$ from $y_0$ and $v_0$ (lesson 6.2.2). In the complex case, remember that $e^{\alpha t}$ adds to the slope.
Raise ValueError if a is 0: the equation is then first order.

The tests use the examples of both lessons, an equation with $a \neq 1$ in each real case, and a growing solution. They compare your answer with solve_ivp for other coefficients, and check that it starts at $y_0$ with slope $v_0$ .

See the tests

import math

import numpy as np
from scipy.integrate import solve_ivp


def _numerical(a, b, c, y0, v0, T):
    sol = solve_ivp(lambda t, s: [s[1], -(b * s[1] + c * s[0]) / a], (0, T), [y0, v0], rtol=1e-12, atol=1e-12)
    return sol.y[0, -1]


def test_two_real_roots():
    """y'' + 5y' + 4y = 0 from y(0) = 1, y'(0) = 0 is 0.484401 at t = 1"""
    expect_close(ivp_solution(1, 5, 4, 1, 0, 1), 0.4844007085990117, what="ivp_solution(1, 5, 4, 1, 0, 1)", rel=1e-9)


def test_repeated_root():
    """y'' + 4y' + 4y = 0 from y(0) = 1, y'(0) = 0 is 3e⁻² at t = 1"""
    expect_close(ivp_solution(1, 4, 4, 1, 0, 1), 3 * math.exp(-2), what="ivp_solution(1, 4, 4, 1, 0, 1)", rel=1e-9)


def test_complex_roots():
    """y'' + 2y' + 5y = 0 from y(0) = 1, y'(0) = 0 is 0.014164 at t = 1"""
    expect_close(ivp_solution(1, 2, 5, 1, 0, 1), 0.01416404894540483, what="ivp_solution(1, 2, 5, 1, 0, 1)", rel=1e-9)


def test_waves():
    """y'' + y = 0 from y(0) = 0, y'(0) = 1 is sin t: sin 2 at t = 2"""
    expect_close(ivp_solution(1, 0, 1, 0, 1, 2), math.sin(2), what="ivp_solution(1, 0, 1, 0, 1, 2)", rel=1e-9)


def test_growth():
    """y'' − y = 0 from y(0) = 1, y'(0) = 0 is (e + 1/e)/2 at t = 1"""
    expect_close(ivp_solution(1, 0, -1, 1, 0, 1), (math.e + 1 / math.e) / 2, what="ivp_solution(1, 0, -1, 1, 0, 1)", rel=1e-9)


def test_leading_coefficient():
    """2y'' + 3y' + y = 0 from y(0) = 1, y'(0) = 1 is −3e⁻² + 4e⁻¹ at t = 2"""
    expect_close(ivp_solution(2, 3, 1, 1, 1, 2), -3 * math.exp(-2) + 4 * math.exp(-1), what="ivp_solution(2, 3, 1, 1, 1, 2)", rel=1e-9)


def test_repeated_with_a():
    """0.5y'' + 2y' + 2y = 0 has a repeated root −2: from y(0) = 1, y'(0) = −1 it is 4e⁻⁶ at t = 3"""
    expect_close(ivp_solution(0.5, 2, 2, 1, -1, 3), 4 * math.exp(-6), what="ivp_solution(0.5, 2, 2, 1, -1, 3)", rel=1e-9)


def test_against_solve_ivp():
    """It agrees with solve_ivp for y'' + 0.3y' + 2.5y = 0 and 3y'' − y' − 2y = 0"""
    for args, T in [((1, 0.3, 2.5, -1, 2), 4), ((3, -1, -2, 0.5, 0.5), 2)]:
        expect_close(ivp_solution(*args, T), _numerical(*args, T), what=f"ivp_solution{args + (T,)}", rel=1e-7)


def test_starts_right():
    """The solution of y'' + y' + 3y = 0 starts at y(0) = 2 with slope −1"""
    h = 1e-6
    expect_close(ivp_solution(1, 1, 3, 2, -1, 0), 2.0, what="y(0)", rel=1e-12)
    slope = (ivp_solution(1, 1, 3, 2, -1, h) - ivp_solution(1, 1, 3, 2, -1, -h)) / (2 * h)
    expect_close(slope, -1.0, what="slope at 0", rel=1e-6)


def test_array():
    """An array of times gives an array of values"""
    t = np.linspace(0, 3, 7)
    got = ivp_solution(1, 2, 5, 1, 0, t)
    assert isinstance(got, np.ndarray) and got.shape == t.shape, f"expected an array of shape {t.shape}, got {got!r}"
    expect_close(got, np.exp(-t) * (np.cos(2 * t) + 0.5 * np.sin(2 * t)), what="values at 0, 0.5, ..., 3", rel=1e-9)


def test_first_order():
    """a = 0 raises ValueError: there is no second derivative"""
    try:
        ivp_solution(0, 1, 2, 1, 0, 1)
    except ValueError:
        return
    raise AssertionError("expected ValueError for a = 0")

The characteristic equation

General solutions and boundary conditions

Coding exercise

Free further reading