Differential Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 9.2 – Direction Fields and Euler's Method

• Section 9.3 – Separable Equations

• Section 9.4 – Models for Population Growth

• Section 3.8 – Exponential Growth and Decay

Expected Educational Results

• Objective 15–01: I can identify ordinary differential equations (ODEs).

• Objective 15–02: I can identify the order of ODEs.

• Objective 15–03: I can solve first-order separable ordinary differential equations.

• Objective 15–04: I can solve first-order separable initial value problems.

• Objective 15–05: I can use technology to sketch a direction field for a first-order ordinary differential equation.

• Objective 15–06: I can model population growth and decay with a differential equation.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Differential Equations

Initial Value Problems (IVPs)

Definition: Initial Value Problem (IVP)

An initial value problem is a DE with an appropriate number of initial conditions (i.e., points). The IVP results in one (1) solution curve, i.e., no constant of integration.

Solving IVPs

There are two (2) methods to solve IVPs:

1. Analytic Method – using theorems to manipulate symbols to find exact solutions;

2. Numeric Methods – finding an approximate solution;

We will discuss only one (1) analytic method: Solving first-order separable IVPs.

Procedure to Solving First-Order IVPs

Suppose an ODE can be written in the form ${\displaystyle \frac{dy}{dx}=f(x)g(y)}$$\displaystyle \frac{dy}{dx}=f(x)g(y)$, $y(x_0)=y_0$$y(x_0)=y_0$.

2. Then the ODE can be rewritten as: ${\displaystyle \frac{dy}{g(y)}=f(x)dx}$$\displaystyle \frac{dy}{g(y)}=f(x)\,dx$.

3. Integrate both sides of the equation: ${\displaystyle \int \frac{dy}{g(y)}=\int f(x)dx}$$\displaystyle \int{\frac{dy}{g(y)}}=\int{f(x)\,dx}$.

4. Use $+C$$+C$ only with the integral of the independent variable.

5. Enter $x_0$$x_0$ into all $x$$x$-variables of the solution; $y_0$$y_0$ into all $y$$y$-variables of the solution; solve for $C$$C$ [or $k$$k$].

6. Solve for $y$$y$, if possible.

7. Check your answer.

Example 02: Solve ${\displaystyle \frac{dy}{dx}=\frac{y}{x+1}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x+1}$ , $y(0)=3$$y(0)=3$

Solution:

1. From Example 01, the general solution is ${\displaystyle y=k(x+1)}$$\displaystyle y=k(x+1)$

2. $y(0)=1\Rightarrow x_0=0\text{ and }{y}_0=3$$y(0)=1\Rightarrow {\color{red}{x_0=0}} \text{ and } {\color{blue}y_0=3}$, So

   • ${\displaystyle \Rightarrow 3=k(0+1)}$$\displaystyle\Rightarrow {\color{blue}3}=k({\color{red}0}+1)$ ${\displaystyle \Rightarrow k=3}$$\displaystyle\Rightarrow k=3$

3. The solution to the IVP is: $y=3(x+1)$$y=3(x+1)$

4. Check your answer:

   1. We need the derivative, $\frac{dy}{dx}$$\color{blue}\frac{dy}{dx}$: ${\displaystyle \frac{d}{dx}(y)=\frac{d}{dx}[3(x+1)]\Rightarrow \frac{dy}{dx}=3}$$\displaystyle \frac{d}{dx}(y)=\frac{d}{dx}\left[3(x+1)\right]\Rightarrow \color{blue}\frac{dy}{dx}=3$

   2. Substitute $y$$\color{red}y$ and $\frac{dy}{dx}$$\color{blue}\frac{dy}{dx}$ into the original ODE: $\frac{dy}{dx}=\frac{y}{x+1}\Rightarrow 3=\frac{3\cancel{(x+1)}}{\cancel{x+1}}\Rightarrow 3=3$${\color{blue}\frac{dy}{dx}}=\frac{\color{red}y}{x+1}\Rightarrow {\color{blue}3}=\frac{\color{red}3\cancel{(x+1)}}{\cancel{x+1}}\Rightarrow 3=3$ which is a true statement. Thus, the answer is correct.

Use Technology to Verify Solutions to First-order IVPs.

NOTE: The use of technology in this activity is to verify solutions to first-order IVPs. On all Assessments, you must show work demonstrating your understanding of finding solutions to first-order IVPs.

Mathematica:

1
(* Example 03: Solve dy/dx=y/(x+1), y(0)=3 *)
2
DSolve[{y[x]/(x+1),y[0]==3},y[x],x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. Mathematica requires two $==$$==$ for the equals sign in an equation.

3. You may need parens, $($$(\,$ and $)$$\,)$, to group multiple terms or factors in the numerator and denominator.

4. DSolve[ ] has at least three arguments:

   1. The IVP to be solved. Using $\{\text{ and }\}$$\{ \text{ and } \}$ group

      1. The ODE;

      2. The initial condition, $\text{y[0]==3}$$\text{y[0]==3}$

   2. The dependent variable in function notation, i.e., $\text{y[x]}$$\text{y[x]}$;

   3. The independent variable, $x$$x$.

5. For help on using the DSolve[ ] function:

   1. In Mathematica, execute the code: $\text{?DSolve}$$\text{?DSolve}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the DSolve[ ] function – you will be able to copy-paste code.

6. Remember, correct Mathematica code will be all black except for variables.

7. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

8. Mathematica may return a result in a different, but equivalent, form than your answer.

Mathematica solution:

${\displaystyle \{y(x)\to 3(x+1)\}}$$\displaystyle \{y(x) \rightarrow 3 (x+1)\}$

Investigation 04

Solve the following IVPs.

1. ${\displaystyle \frac{dy}{dx}=1-y}$$\displaystyle \frac{dy}{dx}=1-y$, $y(0)=-1$$y(0)=-1$

2. ${\displaystyle \frac{dy}{dx}=x(1-y)}$$\displaystyle \frac{dy}{dx}=x(1-y)$, $y(1)=-3$$y(1)=-3$

3. ${\displaystyle \frac{dy}{dx}=e^{x+y}}$$\displaystyle \frac{dy}{dx}=e^{x+y}$, $y(0)=2$$y(0)=2$

4. ${\displaystyle \frac{dy}{dx}=\frac{y}{x^2+1}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x^2+1}$, $y(0)=1$$y(0)=1$

5. ${\displaystyle \frac{dy}{dx}=\frac{1-y}{1+x}}$$\displaystyle \frac{dy}{dx}=\frac{1-y}{1+x}$, $y(0)=1$$y(0)=1$

6. ${\displaystyle \frac{dy}{dx}=\frac{(1-x^2)y^3}{x^2}}$$\displaystyle \frac{dy}{dx}=\frac{(1-x^2)y^3}{x^2}$, $y(2)=1$$y(2)=1$

7. ${\displaystyle x\frac{dy}{dx}=y-3}$$\displaystyle x\frac{dy}{dx}=y-3$, $y(1)=1$$y(1)=1$

8. ${\displaystyle \frac{dy}{dx}=1+x-y-xy}$$\displaystyle \frac{dy}{dx}=1+x-y-xy$, $y(0)=2$$y(0)=2$

Homework

At this time, you should be able to complete the following assignments:

• Section 9.3: # 13, 15, 17

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Created: Thursday, 15 October 2020 6:42 EDT Last Modified: Thursday, 14 July 2022 - 09:57 (EDT)