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

Solving ODEs

There are mathematics courses designed to discuss solving ODEs, e.g., Math 2652 – Introduction to Differential Equations.

There are three (3) methods to solve ODEs:

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

2. Numeric Methods – finding an approximate solution (for solving IVPs);

3. Graphical Methods – using a graph, called Direction Fields (also, Vector Fields), to understand the properties of solutions.

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

Definition: First-Order Separable ODEs

Procedure to Solving First-Order Separable ODEs

Suppose an ODE can be written (using any valid algebraic procedure) in the form ${\displaystyle \frac{dy}{dx}=f(x)g(y)}$$\displaystyle \frac{dy}{dx}=f(x)g(y)$.

1. Then the ODE can be rewritten (using only multiplication or division) as: ${\displaystyle \frac{dy}{g(y)}=f(x)dx}$$\displaystyle \frac{dy}{g(y)}=f(x)\,dx$. NOTE: The variables in the above equation have been separated, i.e., all $x$$x$ factors are on one side of the ODE and all $y$$y$ factors are on the other side of the ODE. You may need to factor prior to separating the variables.

2. 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}$.

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

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

5. Check your answer.

Example 01: Solve ${\displaystyle \frac{dy}{dx}=\frac{y}{x+1}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x+1}$

Solution:

1. The ODE is a first-order separable ODE because ${\displaystyle \frac{dy}{dx}=y\left(\frac{1}{x+1}\right)}$$\displaystyle \frac{\color{red}dy}{\color{blue}dx}={\color{red}y}\left({\color{blue}\frac{1}{x+1}}\right)$

2. Rewrite:

   1. Multiply both sides of the ODE by the differential $dx$$dx$ ${\displaystyle \cancel{dx}\frac{dy}{\cancel{dx}}=y\left(\frac{1}{x+1}\right)dx}$$\displaystyle \cancel{dx}\frac{dy}{\cancel{dx}}=y\left(\frac{1}{x+1}\right)dx$

   2. Divide both sides of the ODE by $y$$y$ ${\displaystyle \frac{dy}{y}=\cancel{y}\left(\frac{1}{x+1}\right)dx\frac{1}{\cancel{y}}}$$\displaystyle \frac{dy}{y}=\cancel{y}\left(\frac{1}{x+1}\right)dx\frac{1}{\cancel{y}}$

3. Integrate both sides of the ODE: ${\displaystyle \int \frac{dy}{y}=\int \frac{dx}{x+1}\Rightarrow \ln |y|=\ln |x+1|+C}$$\displaystyle \int{\color{red}\frac{dy}{y}}=\int{\color{blue}\frac{dx}{x+1}}\Rightarrow \ln{|\,y\,|}=\ln{|\,x+1\,|}+C$

4. Solve for $y$$y$:

   1. Rewrite answer using the exponential form of the logarithm: ${\displaystyle \ln |y|=\ln |x+1|+C\Rightarrow |y|=e^{\ln |x+1|+C}}$$\displaystyle \ln{|\,y\,|}=\ln{|\,x+1\,|}+C\Rightarrow |\,y\,|=e^{\ln{|\,x+1\,|}+C}$

   2. Use property of exponents to rewrite the right-side of the solution: ${\displaystyle |y|=e^{\ln |x+1|+C}\Rightarrow |y|=e^{\ln |x+1|}{e}^C}$$\displaystyle |\,y\,|=e^{\ln{|\,x+1\,|}+C}\Rightarrow |\,y\,|=e^{\ln{|\,x+1\,|}}e^{C}$

   3. We know $e^C>0$$e^C>0$; however, let ${\displaystyle k=e^C}$$\displaystyle k=e^C$, where $k$$k$ is any Real number (positive, negative, or zero), i.e., $e^C=k\in \mathbb{R}$$e^C=k\in\mathbb{R}$ ${\displaystyle |y|=e^{\ln |x+1|}{e}^C\Rightarrow y=k(x+1)}$$\displaystyle |\,y\,|=e^{\ln{|\,x+1\,|}}e^{C}\Rightarrow y=k(x+1)$

   4. Thus, the solution is: ${\displaystyle y=k(x+1)}$$\displaystyle \color{red} y=k(x+1)$

5. Check your answer:

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

   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 k=\frac{k\cancel{(x+1)}}{\cancel{x+1}}\Rightarrow k=k$${\color{blue}\frac{dy}{dx}}=\frac{\color{red}y}{x+1}\Rightarrow {\color{blue}k}=\frac{\color{red}k\cancel{(x+1)}}{\cancel{x+1}}\Rightarrow k=k$ which is a true statement. Thus, the answer is correct.

Use Technology to Verify Solutions to First-order Separable ODEs.

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

Mathematica:

1
(* Example 02: Solve dy/dx=y/(x+1) *)
2
DSolve[y'[x]==y[x]/(x+1),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. Mathematica does not recognize implicit multiplication, e.g., $xy$$xy$; to multiply $x$$x$ with $y$$y$ in Mathematica, you must use a space, $\text{x}\text{y}$$\text{x\,\,y}$, or an asterisk, $\text{x}\ast \text{y}$$\text{x}\ast\text{y}$.

5. DSolve[ ] has at least three arguments:

   1. The ODE to be solved;

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

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

6. 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.

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

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

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

Mathematica solution:

${\displaystyle \{\{y(x)\to C[1](x+1)\}\}}$$\displaystyle \{\{y(x) \rightarrow C[1] (x+1)\}\}$

which looks similar to our solution for Example 01.

Investigation 03

Solve the following ODEs, if possible.

1. ${\displaystyle \frac{dy}{dx}=\frac{y}{x^2+1}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x^2+1}$

2. ${\displaystyle \frac{dy}{dx}=\frac{x}{\sec (x)\sin (y)}}$$\displaystyle \frac{dy}{dx}=\frac{x}{\sec{(x)}\sin{(y)}}$

3. ${\displaystyle x\frac{dy}{dx}=y-3}$$\displaystyle x\frac{dy}{dx}=y-3$

4. ${\displaystyle \frac{dy}{dx}=1+x-y-xy}$$\displaystyle \frac{dy}{dx}=1+x-y-xy$

5. ${\displaystyle \frac{dy}{dx}={\cos }^2(y)}$$\displaystyle \frac{dy}{dx}=\cos^2{(y)}$

6. ${\displaystyle \frac{dy}{dx}=\frac{y}{x}}$$\displaystyle \frac{dy}{dx}=\frac{y}{x}$

7. ${\displaystyle \frac{dy}{dx}=\frac{x}{y}}$$\displaystyle \frac{dy}{dx}=\frac{x}{y}$

8. ${\displaystyle \frac{dy}{dx}=3x^2{e}^{-y}}$$\displaystyle \frac{dy}{dx}=3x^2e^{-y}$

9. ${\displaystyle x{\sin }^2(y)\frac{dy}{dx}=x^2+1}$$\displaystyle x\sin^2{(y)}\frac{dy}{dx}=x^2+1$

10. ${\displaystyle y^{\prime }=e^{2x+y}}$$\displaystyle y^{\prime}=e^{2x+y}$

11. ${\displaystyle y^{\prime }=\frac{y+1}{x-1}}$$\displaystyle y^{\prime}=\frac{y+1}{x-1}$

12. ${\displaystyle y^{\prime }=\frac{xy}{x+1}}$$\displaystyle y^{\prime}=\frac{xy}{x+1}$

13. ${\displaystyle y^{\prime }=\frac{xy+2y-x-2}{xy-3y+x-3}}$$\displaystyle y^{\prime}=\frac{xy+2y-x-2}{xy-3y+x-3}$

Use Technology to Factor Expressions.

NOTE: On all Assessments, you may use technology to factor expressions without showing any work.

Mathematica:

xxxxxxxxxx
2
1
(* Investigation 03-13: Factor (xy+2y-x-2)/(xy-3y+x-3) *)
2
Factor[(x*y+2y-x-2)/(x*y-3y+x-3)]

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

3. Mathematica does not recognize implicit multiplication, e.g., $xy$$xy$; to multiply $x$$x$ with $y$$y$ in Mathematica, you must use a space, $\text{x}\text{y}$$\text{x\,\,y}$, or an asterisk, $\text{x}\ast \text{y}$$\text{x}\ast\text{y}$.

4. Factor[ ] has one argument:

   1. The expression to be factored.

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

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

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

   3. Click on local

   4. Read how to use the Factor[ ] 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.

Homework

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

• Section 9.3: # 1, 3, 5, 7, 9, 11

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Created: Thursday, 15 October 2020 6:42 EDT Last Modified: Thursday, 14 July 2022 - 09:30 (EDT)