1.10 Exercises

These are homework exercises to accompany Libl's "Differential Equations for Engineering" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.

1.1: Integrals as solutions

1.2: Slope fields

Your Turn

Match equations $y^{'} = \sin x$ , $y^{'} = \cos y$ , $y^{'} = y \cos (x)$ to slope fields. Justify.

Answer

$y^{'} = \cos y$
$y^{'} = y \cos (x)$
$y^{'} = \sin x$

Justification left to reader.

Your Turn

(tricky)

Suppose

$0 i f y greater than 0,; 1 i f y less than or equal to 0.$

Does $y^{'} = f (y)$ , $y open parenthesis 0 close parenthesis equals 0$ have a continuously differentiable solution? Does Picard apply? Why, or why not?

Answer

Picard does not apply as $f$ is not continuous at $y equals 0$ . The equation does not have a continuously differentiable solution. Suppose it did. Notice that $y prime open parenthesis 0 close parenthesis equals 1$ . By the first derivative test, $y open parenthesis x close parenthesis greater than 0$ for small positive $x$ . But then for those $x$ we would have $y prime open parenthesis x close parenthesis equals 0$ , so clearly the derivative cannot be continuous.

1.3: Separable Equations

1.4: Linear equations and the integrating factor

In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot be found. If you can find a closed form solution, you should give that.

Your Turn

Suppose we have bacteria on a plate and suppose that we are slowly adding a toxic substance such that the rate of growth is slowing down. That is, suppose that $\frac{d P}{d t} = (2 - 0.1 t) P$ . If $P open parenthesis 0 close parenthesis equals 1000$ , find the population at $t equals 5$ .

Answer

$P (5) = 1000 e^{2 \times 5 - 0.05 \times 5^{2}} = 1000 e^{8.75} \approx 6.31 \times 10^{6}$

1.5: Substitution

Hint: Answers need not always be in closed form.

1.6: Autonomous equations

Your Turn

Assume that a population of fish in a lake satisfies $\frac{d x}{d t} = k x (M - x)$ . Now suppose that fish are continually added at $A$ fish per unit of time.

Find the differential equation for $x$ .
What is the new limiting population?

Answer

$\frac{d x}{d t} = k x (M - x) + A$
$\frac{k M + \sqrt{(k M)^{2} + 4 A k}}{2 k}$

1.7: Numerical methods: Euler’s method

Your Turn

Example of numerical instability: Take $y^{'} = - 5 y$ , $y open parenthesis 0 close parenthesis equals 1$ . We know that the solution should decay to zero as $x$ grows. Using Euler’s method, start with $h equals 1$ and compute $y sub 1, y sub 2, y sub 3, y sub 4$ to try to approximate $y open parenthesis 4 close parenthesis$ . What happened? Now halve the interval. Keep halving the interval and approximating $y open parenthesis 4 close parenthesis$ until the numbers you are getting start to stabilize (that is, until they start going towards zero). Note: You might want to use a calculator.

The simplest method used in practice is the Runge-Kutta method. Consider $\frac{d y}{d x} = f (x, y)$ , $y open parenthesis x sub 0 close parenthesis equals y sub 0$ and a step size $h$ . Everything is the same as in Euler’s method, except the computation of $y sub i plus 1$ and $x sub i plus 1$ .

$\begin{matrix} k_{1} = f (x_{i}, y_{i}), \\ k_{2} = f (x_{i} + \frac{h}{2}, y_{i} + k_{1} \frac{h}{2}) x_{i + 1} = x_{i} + h, \\ k_{3} = f (x_{i} + \frac{h}{2}, y_{i} + k_{2} \frac{h}{2}) y_{i + 1} = y_{i} + \frac{k_{1} + 2 k_{2} + 2 k_{3} + k_{4}}{6} h, \\ k_{4} = f (x_{i} + h, y_{i} + k_{3} h) . \end{matrix}$

There is a simple way to improve Euler’s method to make it a second order method by doing just one extra step. Consider $\frac{d y}{d x} = f (x, y)$ , $y open parenthesis x sub 0 close parenthesis equals y sub 0$ , and a step size $h$ . What we do is to pretend we compute the next step as in Euler, that is, we start with $(x_{i}, y_{i})$ , we compute a slope $k sub 1 equals f open parenthesis x sub i, y sub i close parenthesis$ , and then look at the point $(x_{i} + h, y_{i} + k_{1} h)$ . Instead of letting our new point be $(x_{i} + h, y_{i} + k_{1} h)$ , we compute the slope at that point, call it $k sub 2$ , and then take the average of $k sub 1$ and $k sub 2$ , hoping that the average is going to be closer to the actual slope on the interval from $x_{i}$ to $x_{i} + h$ . And we are correct, if we halve the step, the error should go down by a factor of $2 squared equals 4$ . To summarize, the setup is the same as for regular Euler, except the computation of $y sub i plus 1$ and $x sub i plus 1$ .

$\begin{matrix} k_{1} = f (x_{i}, y_{i}), x_{i + 1} = x_{i} + h, \\ k_{2} = f (x_{i} + h, y_{i} + k_{1} h), y_{i + 1} = y_{i} + \frac{k_{1} + k_{2}}{2} h . \end{matrix}$

1.8 Exact Equations

Your Turn

Solve the following exact equations, implicit general solutions will suffice:

$(2 x y + x^{2}) d x + (x^{2} + y^{2} + 1) d y = 0$
$x^{5} + y^{5} \frac{d y}{d x} = 0$
$e^{x} + y^{3} + 3 x y^{2} \frac{d y}{d x} = 0$
$(x + y) \cos (x) + \sin (x) + \sin (x) y^{'} = 0$

Your Turn

Find the integrating factor for the following equations making them into exact equations:

$e^{x y} d x + \frac{y}{x} e^{x y} d y = 0$
$\frac{e^{x} + y^{3}}{y^{2}} d x + 3 x d y = 0$
$4 (y^{2} + x) d x + \frac{2 x + 2 y^{2}}{y} d y = 0$
$2 \sin (y) d x + x \cos (y) d y = 0$

Your Turn

A function $u (x, y)$ is said to be a if $u_{x x} + u_{y y} = 0$ .

Show if $u$ is harmonic, $- u_{y} d x + u_{x} d y = 0$ is an exact equation. So there exists (at least locally) the so-called function $v (x, y)$ such that $v_{x} = - u_{y}$ and $v_{y} = u_{x}$ .

Verify that the following $u$ are harmonic and find the corresponding harmonic conjugates $v$ :

$u equals 2 x y$
$u = e^{x} \cos y$
$u = x^{3} - 3 x y^{2}$

Your Turn

Solve the following exact equations, implicit general solutions will suffice:

$\cos (x) + y e^{x y} + x e^{x y} y^{'} = 0$
$(2 x + y) d x + (x - 4 y) d y = 0$
$e^{x} + e^{y} \frac{d y}{d x} = 0$
$(3 x^{2} + 3 y) d x + (3 y^{2} + 3 x) d y = 0$

Answer

$e^{x y} + \sin (x) = C$
$x^{2} + x y - 2 y^{2} = C$
$e^{x} + e^{y} = C$
$x^{3} + 3 x y + y^{3} = C$

Your Turn

Find the integrating factor for the following equations making them into exact equations:

$\frac{1}{y} d x + 3 y d y = 0$
$d x - e^{- x - y} d y = 0$
$(\frac{\cos (x)}{y^{2}} + \frac{1}{y}) d x + \frac{x}{y^{2}} d y = 0$
$(2 y + \frac{y^{2}}{x}) d x + (2 y + x) d y = 0$

Answer

Integrating factor is $y$ , equation becomes $d x + 3 y^{2} d y = 0$ .
Integrating factor is $e^{x}$ , equation becomes $e^{x} d x - e^{- y} d y = 0$ .
Integrating factor is $y^{2}$ , equation becomes $(\cos (x) + y) d x + x d y = 0$ .
Integrating factor is $x$ , equation becomes $(2 x y + y^{2}) d x + (x^{2} + 2 x y) d y = 0$ .

1.9: First Order Linear PDE

Your Turn

Solve

$u_{t} - 5 u_{x} = 0$ , $u (x, 0) = \frac{1}{1 + x^{2}}$ ,
$u_{t} + 2 u_{x} = 0$ , $u open parenthesis x, 0 close parenthesis equals cos open parenthesis x close parenthesis$ .

Answer

$u = \frac{1}{1 + (x + 5 t)^{2}}$
$u = \cos (x - 2 t)$

Adapted from Differential Equations for Engineers by Jiří Lebl (Oklahoma State University), hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.