# Topic 2 Introduction to Differential Equations

## 2.1 Basic concepts

### 2.1.1 How to define a differential equation in Maple

Depending on the differentiation command, there are three ways to define a differential equation in Maple.

**Example 2.1 **Assign the differential equation \(y'=2y\) to a variable in Maple.

*Solution*.Method 1: Using the

`diff`

command`ode111:=diff(y(x), x)=2*y(x)`

Method 2: Using the the prime derivative notation

`ode112:=y'=2y`

Method 3: Using the command

`D`

`ode113:=D(y)(x)=2*y(x)`

Among the three methods, the one using `diff`

is the standard choice.

**Exercise 2.1 **Assign the differential equation \(y'=y(1-0.001y)\) to a variable in Maple.

### 2.1.2 How to check solutions

To check if a function (explicitly of implicitly defined) is a solution of a given differential equation, you may use the `odetest(function, ODE, y(x))`

command. If the output is 0, then the function is a solution.

**Example 2.2 **Verify that \(y=ce^{2x}\) is a solution to the differential equation \(y'=2y\).

*Solution*. Run the following command, you will see that output is 0. So the function is a solution.

`odetest(y(x)=c*exp(2*x), diff(y(x), x)=2*y(x), y(x))`

**Example 2.3 **Verify that \(x^2+y^2=1\) is a solution of the equation \(y'=-\frac xy\).

*Solution*. Again, running the following command returns the number 0. So the implicit function is a solution.

`odetest(x^2+y(x)^2=1, diff(y(x),x)=-x/(y(x)), y(x));`

When working with differential equations, we should always use \(y(x)\) instead of \(y\) to indicate that \(y\) is a function of \(x\).

**Exercise 2.2 **Verify that the function \(y=c_1e^x+c_2e^{-x}\) is a solution of the equation \(y''=y\).

**Exercise 2.3 **Verify that \(x^2+4y^2=c\) is an implicit solution of the equation \(4yy'=x\).

## 2.2 Solution curves vs Integral curves

A solution curve is the graph of a function \(y=f(x)\) that satisfies the given differential equation. An integral curves is a union of solution curves.

**Example 2.4 **Consider the differential equation \(yy'=4x\). Verify that the graphs of the functions \(y=\pm\sqrt{4x^2-1}\) are solution curves of the equation, while the hyperbola \(4x^2-y^2=1\) defines an integral curve of the equation.

*Solution*. Let’s rename the function as \(y_1(x)=\sqrt{4x^2-1}\) and \(y_2(x)=-\sqrt{4x^2-1}\). In Maple, it means we `y[1](x)`

and `y[2](x)`

. We can check that they are solutions using the `seq`

loop.

```
ode121:=y(x)*diff(y(x), x)=4*x:
y[1](x):=sqrt(4*x^2-1);
y[2](x):=-sqrt(4*x^2-1);
seq(odetest(y(x)=y[i](x), ode121, y(x)), i = 1 .. 2);
```

The outputs show that \(y_1\) and \(y_2\) are solutions.

To see that hyperbola \(4x^2-y^2=1\) defines an integral curve, we solve for \(y\).

`soly:=solve(4*x^2-y^2=1, y);`

You will see that the solutions are exactly the functions \(y_1\) and \(y_2\). So as an union of solution curves the hyperbola is an integral curve.

Plotting those curves will help use understand better.

```
solutioncurves := plot([y[1](x), y[2](x)],
x = -5 .. 5, y = -5 .. 5, color = [green, red]);
with(plots):
integralcurve := implicitplot(4*x^2 - y^2 = 1,
x = -5 .. 5, y = -5 .. 5, color = blue,
linestyle = dot);
```

To check the integral curve is the union of the two solution curve, we can use the `display`

command.

`display(solutioncurves,integralcurve)`

**Exercise 2.4 **Consider the differential equation \(yy'=4x\). Verify that the graphs of the functions \(y=\pm\sqrt{4x^2+1}\) are solution curves of the equation, while the hyperbola \(y^2-4x^2=1\) defines an integral curve of the equation.

## 2.3 Direction fields

In Maple, the commands `DEplot`

, `dfieldplot`

, and `phaseportrait`

supported by the package `DETools`

can be used to plot the direction field and solution curves. The basic usage is as follows

`DEplot(differential equation, function, ranges, options)`

Again, using the command `?DEplot`

, we can find details and examples on the command. In the following, I will use `DEplot`

as an example to show how they work.

**Example 2.5 **Plot the direction field for the differential equation \(y'=-\frac xy\). Can you guess what is a solution to this equation?

*Solution*. Let’s first define the differential equations and assign it to a variable.

`ode131:=diff(y(x), x)=-x/(y(x));`

Now load the package `DETools`

using the command `with()`

.

`with(DETools):`

With the package loaded, we can use `DEplot`

to plot the direction field for `ode131`

, say in the region \(-5\le x\le 5\) and \(-5\le y\le 5\).

```
DEplot(ode131, y(x), x =-5..5, y =-5..5,
title = " Direction Field for y'=-x/y ")
```

Note that one may change the outlook by add options, such as color and arrows. Running the following command, you will see the difference.

```
DEplot(diff(y(x), x) = -(x - 1)/(y(x) + 1), y(x),
x = -5 .. 5, y = -5 .. 5,
title = "Direction Field for y'=2 y",
color = -(x - 1)/(y(x) + 1), arrows = line)
```

The direction field suggests that solutions are circles.

To display \(y(x)\) as \(y\) in the output, you may run the following commands first.

```
with(PDEtools, declare):
declare(y(x), prime=x); # Turn ON the enhanced DEdisplay feature
```

You will see the following output \[ {\color{blue} \text{derivatives with respect to } x \text{ of functions of one variable will now be displayed with '} } \]

**Exercise 2.5 **Plot the direction field for the differential equation \(y'=-\frac{x-1}{y+1}\). Can you guess what is a solution to this equation?

**Exercise 2.6 **Plot the direction field for the differential equation \(y'=-\frac{x}{2y}\). Can you guess what is a solution to this equation?