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The Ordinary Differential Equations Project

Thomas W. Judson

Appendix B Hints and Answers to Selected Exercises

1 A First Look at Differential Equations
1.1 Modeling with Differential Equations
1.1.9 Exercises

1.1.9.29.

Hint.
Rewriting the differential equation as \(x' + a x - q(t) = 0\) and using the fact that
\begin{equation*} x'(t) = -a Ce^{-at} - a e^{-at} \int_{t_0}^t e^{as}q(s) \, ds + b(t), \end{equation*}
we see that
\begin{align*} x'(t) + a x(t) - q(t) & = -a Ce^{-at} - a e^{-at} \int_{t_0}^t e^{as}q(s) \, ds + q(t)\\ & + \; aCe^{-at} + ae^{-at} \int_{t_0}^t e^{as}q(s) \, ds - q(t)\\ & = 0. \end{align*}

1.1.9.31.

Hint.
Think about the limit of the interaction term as the number of prey becomes very large.

1.4 Analyzing Equations Numerically
1.4.6 Exercises

1.4.6.5.

Hint.
This equation is a first-order linear equation (Section 1.5), but it is possible to find the analytic solution using Sage (Subsection 1.2.10).

1.4.6.8.

Hint.
Hints for part (2):
  • For fixed \(i\) show that
    \begin{align*} a_{i + 1} & \leq (1 + s)a_i + t\\ & \leq (1 + s)[(1 + s)a_{i - 1}+ t] + t\\ & \leq (1 + s)\{ (1 + s)[(1 + s) a_{i - 2} + t]+ t\} + t\\ & \vdots\\ & \leq (1 + s)^{i+1}a_0 + [1 + (1 + s) + (1+s)^2 + \cdots + (1 + s)^i]t. \end{align*}
  • Now use a geometric sum to show that
    \begin{equation*} a_{i + 1} \leq (1 + s)^{i + 1} a_0 + \frac{t}{s}[(1 + s)^{i + 1} - 1] = (1 + s)^{i + 1} \left( \frac{t}{s} + a_0 \right) - \frac{t}{s}. \end{equation*}
  • Apply part (1) to derive
    \begin{equation*} a_{i + 1} \leq e^{(i + 1)s} \left( \frac{t}{s} + a_0 \right) - \frac{t}{s}. \end{equation*}

1.5 First-Order Linear Equations
1.5.8 Exercises

1.5.8.16.

Hint.
\(y = \dfrac{3 x^{4} + 8 x^{3} + 6 x^{2} + 12}{6 (x + 1)^2}\)

1.5.8.17.

Hint.
\(y = (e^x + 1)/(x+1)^2\)

1.5.8.19.

Hint.
\(y = e^x \sec x\)

1.5.8.20.

Hint.
\(y = \sin x / (x + 2)\)

1.5.8.21.

Hint.
If \(x(t)\) is the amount of salt in the tank at time \(t\text{,}\) we know that \(x(0) = 10\text{.}\) The volume of the tank is \(V = 200 + 5t\text{.}\) We can model the amount of salt in the tank at time \(t\) with a differential equation,
\begin{align*} \frac{dx}{dt} & = \text{rate in} - \text{rate out}\\ & = 10(0.1) - 5 \frac{x}{V}\\ & = 1 - 5\frac{x}{200+ 5t}\\ & = 1 - \frac{x}{40 + t}. \end{align*}
The resulting equation
\begin{equation*} \frac{dx}{dt} + \frac{1}{40 + t}x = 1 \end{equation*}
is a first order linear differential equation. An integrating factor for this equation is given by
\begin{equation*} \mu(t) = \exp\left(\int \frac{1}{40 + t} \, dt\right) = 40 + t. \end{equation*}
Multiplying both sides of the differential equation by \(\mu(t)\text{,}\) we have
\begin{equation*} \frac{d}{dt} [(40 + t)x] = (40 + t) \frac{dx}{dt} + x = (40 + t)\left( \frac{dx}{dt} + \frac{1}{40 + t}x \right) = 40 + t. \end{equation*}
Integrating both sides of this equation, we obtain
\begin{equation*} (40 + t)x = 40t + \frac{t^2}{2} + C. \end{equation*}
Using the intial condition \(x(0) = 10\text{,}\) we can determine that \(C = 400\) or
\begin{equation*} x(t) = \frac{t^2 + 80t + 800}{2t + 80}. \end{equation*}
The tank is full at time \(t = 400/5 = 80\text{,}\) and the tank contains \(x(80) = 170/3 \approx 56.67\) kilograms of salt when the tank is full.

1.5.8.27.

Hint.
  1. If \(y = y_1 + 1/v\text{,}\) then \(y' = y_1' - v'/v^2\text{.}\) Substituting into our original equation, we obtain
    \begin{equation*} y' = y_1' - \frac{v'}{v^2} = p + q y_1 + r y_1^2 - \frac{v'}{v^2}. \end{equation*}
    On the other hand,
    \begin{align*} y' & = p + q \left(y_1 + \frac{1}{v} \right) + r \left(y_1 + \frac{1}{v} \right)^2\\ & = p + q y_1 + \frac{q}{v} + r y_1^2 + \frac{2r y_1}{v} + \frac{r}{v^2}\\ & = y_1' + \frac{q}{v} + \frac{2r y_1}{v} + \frac{r}{v^2}. \end{align*}
    Therefore,
    \begin{equation*} - \frac{v'}{v^2} = \frac{q}{v} + \frac{2r y_1}{v} + \frac{r}{v^2}, \end{equation*}
    which is just the first-order linear equation
    \begin{equation*} v' + [q(t) + 2 r(t) y_1(t)]v = - r(t). \end{equation*}
  2. \begin{equation*} y = t + \frac{1}{C - t} \end{equation*}
  3. \begin{equation*} y(t) = \frac{1}{C \cos t - \sin t} + \sin t \end{equation*}
  4. \begin{equation*} y(t) = 2 + \frac{1}{C e^t - 1} \end{equation*}

1.6 Existence and Uniqueness of Solutions
1.6.5 Exercises

1.6.5.1.

Hint.
  1. There exists a unique solution to \(y' = y^2 + y^3\text{,}\) \(y(0) = 1\text{,}\) since \(f(t, y) = y^2 + y^3\) and \(\partial f(t, y)/\partial y = 2y + 3y^2\) are continuous at the point \((0, 1)\text{.}\)
  2. The Existence and Uniqueness Theorem does not apply to \(y' = \sqrt[4]{y}\text{,}\) \(y(1) = 0\text{,}\) since \(f(t, y) = \sqrt[4]{y}\) is not continuous at \((1, 0)\text{.}\)
  3. There exists a unique solution to \(y' = \sqrt[4]{y}\text{,}\) \(y(1) = 1\text{,}\) since \(f(t, y) = \sqrt[4]{y}\) and \(\partial f(t, y)/\partial y = y^{-3/4}/4\) are both continuous at the point \((1, 1)\text{.}\)
  4. The Existence and Uniqueness Theorem does not apply to \(x' = t/(x^2 - 4)\text{,}\) \(x(0) = 2\text{,}\) since \(f(t, x) = t/(x^2 - 4)\) is not continuous at \((0, 2)\text{.}\)
  5. There exists a unique solution to \(x' = t/(x^2 - 4)\text{,}\) \(x(2) = 0\text{,}\) since \(f(t, x) = t/(x^2 - 4)\) and \(\partial f(t, x)/\partial x = - 2tx/(x^2 - 4)^2\) are both continuous at the point \((2, 0)\text{.}\)
  6. There exists a unique solution to \(y' = x \sin y\text{,}\) \(y(0) = 0\text{,}\) since \(f(x, y) = x \sin y\) and \(\partial f(x, y)/\partial y = x \cos y\) are both continuous at the point \((0, 0)\text{.}\)
  7. The Existence and Uniqueness Theorem does not apply to \(y' = 1/(t - 1) y + 2t\text{,}\) \(y(1) = 1\text{,}\) since \(f(t, y) = 1/(t - 1) y + 2t\) is not continuous at \((1, 1)\text{.}\)

1.6.5.3.

Hint.
(b) Make sure that the derivative of \(y(t)\) exists at \(t = t_0\text{.}\)

3 Linear Systems
3.2 Planar Systems
3.2.6 Exercises

3.2.6.10.

Hint.
Assume that your solution must be of the form
\begin{equation*} {\mathbf x}_p = \begin{pmatrix} a_2 t^2 + a_1 t + a_0 \\ b_2 t^2 + b_1 t + b_0. \end{pmatrix} \end{equation*}
This is called the method of undetermined coefficients.

4 Second-Order Linear Equations
4.1 Homogeneous Linear Equations
4.1.6 Exercises

4.1.6.31.

Hint.
Pay careful attention to units.

4.1.6.32.

Hint.
  1. Observe that
    \begin{align*} a x_1'' + b_1' + cx_1 & = a \left(\frac{-b}{2a}\right)^2e^{-bt/2a} + b \left( \frac{-b}{2a} \right) e^{-bt/2a} + c e^{-bt/2a}\\ & = e^{-bt/2a} \left( \frac{b^2}{4a} - \frac{b^2}{2a} + c \right)\\ & = e^{-bt/2a} \left( \frac{-b^2 + 4ac}{4a} \right)\\ & = 0. \end{align*}
  2. If \(y = v(t) x_1(t) = v(t) e^{-bt/2a}\) is a solution to our differential equation, then
    \begin{align*} a y'' + b y' + cy & = a (v''x_1 + 2 v'x_1' + vx_1'' ) + b(v' x_1 + v x_1') + cv x_1\\ & = a v''x_1 + 2a v'x_1' + bv' x_1 + v(a x_1'' +b x_1' + c x_1)\\ & = a v'' e^{-bt/2a} + \left[2a \left( \frac{-b}{2a} \right) e^{-bt/2a} + b e^{-bt/2a} \right] v'\\ & = a v'' e^{-bt/2a}\\ & = 0. \end{align*}
    Since \(a \neq 0\text{,}\) we know that \(v'' = 0\text{.}\) Hence, \(v(t) = c_1 + c_2 t\text{.}\)

4.1.6.33.

Hint.
  1. \begin{align*} x'' + px' + qx & = (v''x_1 + 2 v' x_1' + vx_1'') + p(v'x_1 + vx_1') + q(vx_1)\\ & = x_1 v'' + 2v' x_1' + p x_1 v' + v (x_1'' + p x_1' + q x_1)\\ & = x_1 v'' +(2x_1' + px_1)v' \\ & = 0. \end{align*}
  2. If \(u = v'\text{,}\) then \(x_1 u' +(2x_1' + px_1)u= 0\text{.}\)
  3. If \(x_1(t) = 1/t\text{,}\) then
    \begin{equation*} 2 t^2 x_1'' + 3t x_1' - x_1 = 2 t^2 \left(\frac{2}{t^3}\right) + 3t \left(\frac{-1}{t^2}\right) - \frac{1}{t} = 0. \end{equation*}
    If we assume that \(x = v/t\) is a second solution, then
    \begin{equation*} 2 t^2 x'' + 3t x' - x = 2tv'' - v' = 0. \end{equation*}
    If we let \(u = v'\text{,}\) then a solution of \(2tu' - u = 0\) is \(u = \sqrt{t}\) and \(v = \int \sqrt{t} \, dt = 2 t^{3/2} / 3\text{.}\) Therefore, the second solution to our equation is
    \begin{equation*} x = \frac{v}{t} = \frac{2}{3} \sqrt{t}. \end{equation*}

4.1.6.38.

Hint.
Show that
\begin{align*} \frac{dy}{dt} & = \frac{dx}{dt} \frac{dy}{dx}\\ \frac{d^2y}{dt^2} & = \left( \frac{dx}{dt} \right)^2 \frac{d^2y}{dx^2} + \frac{d^2x}{dt^2} \frac{dy}{dx}. \end{align*}

4.2 Forcing
4.2.7 Exercises

4.2.7.25.

Hint.
Suppose that that \(f(t)\) and \(g(t)\) are linearly dependent on an interval \(I = (a, b)\text{.}\) Then one function is a multiple of the other, say \(f(t) = c g(t)\text{.}\) Thus, \(f'(t) = cg'(t)\text{.}\)
\begin{equation*} W(f, g)(t) = \det \begin{pmatrix} f(t) & g(t) \\ f'(t) & g'(t) \end{pmatrix} = f(t) g'(t) - f'(t) g(t) = c g(t) g'(t) - cg'(t) g(t) = 0. \end{equation*}
Conversely, suppose that
\begin{equation*} W(f, g)(t) = \det \begin{pmatrix} f(t) & g(t) \\ f'(t) & g'(t) \end{pmatrix} = 0, \end{equation*}
for all \(t\) in \((a, b)\text{.}\) If \(g = 0\text{,}\) then \(0 f = g\) and the two functions are linearly dependent. Assume that \(g(t_0) \neq 0\) for some \(t_0\) in \((a, b)\text{.}\) Since \(g\) is differentiable, it must also be continuous and there is some interval \((c, d)\) contained in \((a, b)\) such that \(t_0 \in (c, d)\) and \(g\) does not vanish on this interval. Therefore,
\begin{equation*} \frac{d}{dt} \left( \frac{f}{g} \right) = \frac{f' g - f g'}{g^2} = - \frac{W(f,g)}{g^2} = 0, \end{equation*}
and \(f/g\) is constant on the interval \((c, d)\text{.}\) Thus, \(f(t_0) = c g(t_0)\) and \(f'(t_0) = c g'(t_0)\text{.}\) Since \(f\) and \(cg\) are both solutions to the differential equation \(y'' + p y' + q y = 0\) and have the same initial condition, \(f(t) = cg(t)\) for all \(t \in (a, b)\) by the existence and uniqueness theorem. Consequently, \(f\) and \(g\) are linearly dependent.

4.2.7.26.

Hint.
  1. We can rewrite \(2 t^2 y'' + 3ty' - y = 0\) as
    \begin{equation*} y'' + \frac{3}{2t} y' - \frac{1}{2t^2} y = 0. \end{equation*}
    Since \(p(t) = 1/2t\text{,}\) Abel’s Theorem tells us that
    \begin{equation*} W[y_1, y_2](t) = c \exp\left( - \int \frac{3}{2t} \, dt \right) = c \exp\left( - \frac{3}{2} \ln t \right) = ct^{-3/2}. \end{equation*}
  2. Since \(y_1\) and \(y_2\) are solutions to our differential equation, we know that
    \begin{gather*} y_1'' + p(t) y_1' + q(t) y_1 = 0\\ y_2'' + p(t) y_2' + q(t) y_2 = 0. \end{gather*}
    Multiplying the first equation by \(y_2\) and the second equation by \(y_1\) and subtracting, we obtain
    \begin{equation} (y_1 y_2'' - y_1'' y_2) + p(t) (y_1 y_2' - y_1' y_2) = 0.\tag{4.2.5} \end{equation}
    If
    \begin{equation*} W(t) = W(y_1, y_2)(t) = y_1 y_2' - y_1' y_2, \end{equation*}
    then
    \begin{equation*} W' = y_1 y_2'' - y_1'' y_2, \end{equation*}
    and equation (4.2.5) becomes
    \begin{equation*} W' + p(t) W = 0. \end{equation*}
    This equation is separable with solution
    \begin{equation*} W(t) = c \exp\left( - \int p(t) \, dt \right). \end{equation*}

4.2.7.27.

Hint.
  1. If \(y_p = u_1y_1 + u_2 y_2\text{,}\) then
    \begin{gather*} y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2' = u_1 y_1' + u_2 y_2'\\ y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''. \end{gather*}
    Substituting these expressions into equation (4.2.6), we have
    \begin{align*} y_p'' + p y_p' + q y_p & = ( u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2'') + p(u_1 y_1' + u_2 y_2')\\ & + q(u_1y_1 + u_2 y_2)\\ & = u_1[y_1'' + p y_1' +q y_1] + u_2[y_2'' + p y_2' +q y_2] + u_1' y_1' + u_2' y_2'\\ & = u_1' y_1' + u_2' y_2'\\ & = f(t). \end{align*}
  2. If we solve the system
    \begin{align*} u_1'(t) y_1(t) + u_2'(t) y_2(t) & = 0\\ u_1'(t) y_1' (t)+ u_2'(t) y_2'(t) & = f(t). \end{align*}
    for \(u_1'\) and \(u_2'\text{,}\) we obtain
    \begin{align*} u_1'(t) & = \frac{- y_2(t) f(t)}{W[y_1, y_2](t)}\\ u_2'(t) & = \frac{y_1(t) f(t)}{W[y_1, y_2](t)}. \end{align*}
  3. Integrate the two equations from part (2).
  4. The general solution to the homogeneous equation \(y'' + 4y = 0\) is
    \begin{equation*} y_h = c_1 \cos 2t + c_2 \sin 2t. \end{equation*}
    To find a particular solution, assume that the solution has the form
    \begin{equation*} y_p = u_1(t) \cos 2t + u_2(t) \sin 2t. \end{equation*}
    By part (2)
    \begin{align*} u_1'(t) & = -3 \cos t\\ u_2'(t) & = \frac{3}{2} \csc t - 3 \sin t. \end{align*}
    Integrating, we obtain
    \begin{align*} u_1(t) & = -3 \sin t\\ u_2(t) & = \frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t. \end{align*}
    Therefore,
    \begin{align*} y_p(t) & = u_1(t) y_1(t) + u_2(t) y_2(t)\\ & = -3 \sin t \cos 2t + \left[\frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t\right] \sin 2t, \end{align*}
    and the general solution is
    \begin{align*} y & = y_h + y_p\\ & = c_1 \cos 2t + c_2 \sin 2t -3 \sin t \cos 2t + \left[\frac{3}{2} \ln| \csc t - \cot t| + 3 \cos t\right] \sin 2t. \end{align*}

4.3 Sinusoidal Forcing
4.3.5 Exercises

4.3.5.2.

Hint.
Assume the complex solution has form \(y_c = A e^{3it}\text{.}\)

4.3.5.12.

Hint.
Assume the complex solution has form \(y_c = A e^{3it}\text{.}\)

5 Nonlinear Systems
5.3 More Nonlinear Mechanics
5.3.6 Exercises

5.3.6.17. Lobster Navigation.

Hint.
You may find Sage useful.
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