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

$f(y)=\left\{\begin{array}{cc}{0}&{if\: y>0,} \\ {1}&{if\: y\leq 0.}\end{array}\right.\nonumber$

Does $y'=f(y)$ , $y(0)=0$ have a continuously differentiable solution? Does Picard apply? Why, or why not?

Answer

Picard does not apply as $f$ is not continuous at $y=0$ . The equation does not have a continuously differentiable solution. Suppose it did. Notice that $y'(0)=1$ . By the first derivative test, $y(x)>0$ for small positive $x$ . But then for those $x$ we would have $y'(x)=0$ , 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 $\frac{dP}{dt}=(2-0.1\, t)P$ . If $P(0)=1000$ , find the population at $t=5$ .

Answer

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

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 $\frac{dx}{dt} = kx (M -x)$ . 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

$\frac{dx}{dt}=kx(M-x)+A$
$\frac{kM+\sqrt{(kM)^{2}+4Ak}}{2k}$

1.7: Numerical methods: Euler’s method

Your Turn

Example of numerical instability: Take $y' = -5y$ , $y(0) = 1$ . We know that the solution should decay to zero as $x$ grows. Using Euler’s method, start with $h =1$ and compute $y_1, y_2, y_3, y_4$ to try to approximate $y(4)$ . What happened? Now halve the interval. Keep halving the interval and approximating $y(4)$ 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 $\frac{dy}{dx} = f(x, y)$ , $y(x_0) = y_0$ and a step size $h$ . Everything is the same as in Euler’s method, except the computation of $y_{i+1}$ and $x_{i+1}$ .

$\begin{align}\begin{aligned} k_1 &= f(x_i, y_i), \\ k_2 &= f(x_i + \frac{h}{2}, y_i +k_1 \frac{h}{2}) & x_{i+1} = x_i + h, \\ k_3 &= f(x_i + \frac{h}{2}, y_i +k_2 \frac{h}{2}) & y_{i+1} = y_i + \dfrac{k_1 + 2k_2 + 2k_3 + k_4}{6}h, \\ k_4 &= f(x_i + h, y_i +k_3h).\end{aligned}\end{align}$

There is a simple way to improve Euler’s method to make it a second order method by doing just one extra step. Consider $\frac{dy}{dx}=f(x,y)$ , $y(x_0) = y_0$ , 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 $k_1 = f(x_i,y_i)$ , and then look at the point $(x_i+h,y_i + k_1h)$ . Instead of letting our new point be $(x_i+h,y_i + k_1h)$ , we compute the slope at that point, call it $k_2$ , and then take the average of $k_1$ and $k_2$ , 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 $2^2 = 4$ . To summarize, the setup is the same as for regular Euler, except the computation of $y_{i+1}$ and $x_{i+1}$ .

$\begin{align}\begin{aligned} & k_1 = f(x_i,y_i), & & x_{i+1} = x_i + h, \\ & k_2 = f(x_i + h,y_i + k_1h), & & y_{i+1} = y_i + \frac{k_1+k_2}{2}\,h.\end{aligned}\end{align}$

Your Turn

Consider $\dfrac{dy}{dx} = x+y$ , $y(0)=1$ .

Use the improved Euler’s method (see above) with step sizes $h=\frac{1}{4}$ and $h=\frac{1}{8}$ to approximate $y(1)$ .
Use Euler’s method with $h=\frac{1}{4}$ and $h=\frac{1}{8}$ .
Solve exactly, find the exact value of $y(1)$ .
Compute the errors, and the factors by which the errors changed.

Answer

Improved Euler: $y(1)\approx 3.3897$ for $h=1/4$ , $y(1)\approx 3.4237$ for $h=1/8$ ,
Standard Euler: $y(1)\approx 2.8828$ for $h=1/4$ , $y(1)\approx 3.1316$ for $h=1/8$ ,
$y=2e^{x}-x-1$ , so $y(2)$ is approximately $3.4366$ .
Approximate errors for improved Euler: $0.046852$ for $h=1/4$ , and $0.012881$ for $h=1/8$ . For standard Euler: $0.55375$ for $h=1/4$ , and $0.30499$ for $h=1/8$ . Factor is approximately $0.27$ for improved Euler, and $0.55$ for standard Euler.

1.8 Exact Equations

Your Turn

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

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

Your Turn

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

$e^{xy} \, dx + \frac{y}{x} e^{xy} \, dy = 0$
$\frac{e^x+y^3}{y^2} \, dx + 3x \, dy = 0$
$4(y^2+x) \, dx + \frac{2x+2y^2}{y} \, dy = 0$
$2\sin(y) \, dx + x\cos(y)\, dy = 0$

Your Turn

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

Show if $u$ is harmonic, $-u_y \, dx + u_x \, dy = 0$ 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 $v$ :

$u = 2xy$
$u = e^x \cos y$
$u = x^3-3xy^2$

Your Turn

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

$\cos(x)+ye^{xy} + xe^{xy} y' = 0$
$(2x+y)\, dx + (x-4y) \, dy = 0$
$e^x + e^y \frac{dy}{dx} = 0$
$(3x^2+3y)\,dx + (3y^2+3x)\, dy = 0$

Answer

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

Your Turn

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

$\frac{1}{y}\, dx + 3y \, dy = 0$
$dx - e^{-x-y} \, dy = 0$
$\bigl( \frac{\cos(x)}{y^2} + \frac{1}{y} \bigr) \, dx + \frac{x}{y^2} \, dy = 0$
$\bigl( 2y + \frac{y^2}{x} \bigr) \, dx + ( 2y+x )\, dy = 0$

Answer

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

1.9: First Order Linear PDE

Your Turn

Solve

$u_t - 5u_x = 0$ , $u(x,0) = \frac{1}{1+x^2}$ ,
$u_t + 2u_x = 0$ , $u(x,0) = \cos(x)$ .

Answer

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

Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.