The following example shows how to find a symbolic solution to the ODE $\frac{\ⅆ}{\ⅆ x} y\=a\cdot y$.

Define a variable called basicODE and set it equal to the ODE.

Use dsolve(basicODE) to find the solution. Here, we assign the result to a variable called basicSolution. The result is an equation that satisfies the ODE. Terms of the form _Cn (where n is an integer) are constants of integration.

To find the constant for a particular solution, include an initial value equation with the ODE in a set or list and then pass the set/list to dsolve. The following expression finds a solution that satisfies the condition $y\=5$ when $x\=0$.

Test your solution with the odetest function. The odetest function takes the symbolic solution from dsolve and your ODEs as parameters. It returns 0 if the solution satisfies the ODEs.

To find the value of $y$ at a particular value of $x$ or to graph the solution, you can use either the rhs function or the eval(expression, equations) command to assign the right-hand side of the solution to a variable. Of the two methods, eval is more flexible (especially when dealing with a set of solutions), and will be used in the rest of the examples in this section.

After assigning a value to the parameter $a$, use eval on Y to get an exact value for the solution at a point.

Use the evalf function to get a floating-point value.

To visualize the solution, use the plot function. The following graphs Y over $x=0..5$.

Finally, the function Y can be used in fsolve to find, for example, the value of x for which y(x)=2.

Symbolic solutions can be differentiated and integrated using the commands diff and int.

In this example, we have a first-order differential equation in v(t), the velocity of an object. Once we find v(t), we want to take its derivative to find the object's acceleration, and then take the integral of v(t) to find the distance traveled by the object. Finally, we want to plot the velocity, acceleration, and distance on a single plot.

We start by defining equations, a set that contains the ODE and the initial condition (we assume the initial velocity to be 0).

Run dsolve(equations) to get a solution for v(t).

Assign the solution for v(t) to a variable, V:

You can use the diff function to take the derivative of V and get the acceleration of the object, A.

Similarly, you can use the int command to integrate V and get S, the distance traveled by the object. Since V is an expression in t, we need to integrate V over t from our initial time, 0, to some final time, T, giving us an expression in T. (Note that we assume the initial position of the object to be 0.)

If you are only interested in solutions for T > 0, you can specify this by using assuming on the int expression.

Note that this expression is equal to our initial expression for S, -ln(cosh(T)).

To get an expression in t, evaluate S at t:

The following command plots our solution, V, its derivative, A, and its integral, S.

When solving a system of ODEs, enclose the system in a set or list. The following is an example of a system with two ODEs enclosed in a set.

Run dsolve(odeSystem) to get a solution. The result is a set of equations that satisfies the system.

To access the equations in the solution, use the eval(expression, equations) command. Enter either x(t) or y(t) for expression; enter systemSol for equations.

Include the initial conditions in the set of ODEs to find a particular solution for the system. The following specifies that the solution should go through the point (1,2) at t=0.

Use eval and evalf to analyze your solution.

The following command finds the first time x(t) = y(t) with t ≥ 1.5.

The X and Y variables can be used to graph the solutions using the plot function. The following is a graph of x(t) and y(t) versus t.

The following is a parametric plot of the solution over the same time interval as the preceding graph.

Finding solutions to a two-point boundary value problem (BVP) is more involved than solving an initial value problem. For example, let's say you have an ODE that can always be solved as an initial value problem. When that same ODE is cast as a two-point BVP, there is no guarantee that a solution can be found. In fact, we will see that to a two-point BVP can result in any of the following:

Note: The theory that explains the behavior of the two-point BVP for linear second-order ODEs is attributed to Sturm and Liouville. Such problems are called Sturm-Liouville problems.

Let's start off by finding the solution for an ODE given some initial conditions. After that, we'll look at solving the same ODE but as a two-point BVP.

The general solution has two constants, _C1 and _C2. To get a specific solution, include two initial value equations along with ode in dsolve.

This verifies that we can always find a solution for this ODE given the initial values we defined.

Let's now look at the solutions we get from the same ODE as part of a two-point BVP. First, define two generic boundary value conditions at x = 0 and x = π. Our boundary values are linear combinations of y(x) and y'(x) at the boundaries.

When we solve the system consisting of ode, bc1, and bc2, we get the following. (Note that the solution exists and is unique as long as a⋅d - b⋅c ≠ 0.)

To analyze the solution, assign values to the parameters a, b, c, d, f, and g (ensuring, of course, that a⋅d - b⋅c ≠ 0).

We have found the case where we get a unique solution to the system. To see the other two possibilities (that is, no solution and an infinite number of solutions), let's examine what happens when we set a, b, c, and d such that a⋅d - b⋅c = 0. We do this by setting the boundary values to y(0) = 1 and y(π) = 0.

We get no solution from dsolve. To understand why, recall that the general solution to our ODE is:

y(x) = _C1⋅sin(x) + _C2⋅cos(x) 

When we apply our boundary values to the general solution, we get the following equations for the constants _C1 and _C2.

_C1⋅sin(0) + _C2⋅cos(0) = 1 ⇒ _C2 = 1
_C1⋅sin(π) + _C2⋅cos(π) = 1 ⇒ _C2 = 0

Our boundary values, therefore, do not allow for a solution to our system.

Finally, let's see what happens when we set the boundary values to y(0) = 1 and y(π) = -1.

We get a solution from dsolve, but it does not look complete. One of the constants in the general solution was found, but the other, _C1, remains in the solution. We therefore have infinitely many solutions to this BVP since any multiple of sin(x) can be added to cos(x). To understand why this happens, apply the boundary values to the general solution to get the following equations.

_C1⋅sin(0) + _C2⋅cos(0) = 1 ⇒ _C2 = 1
_C1⋅sin(π) + _C2⋅cos(π) = -1 ⇒ _C2 = 1

We can see that our boundary values enable us to solve for _C2, but give no criteria at all for _C1. This means that _C1 can take on any value, which is consistent with the solution found by dsolve.

The dsolve function always tries to return an explicit solution. If the solution cannot be expressed explicitly, dsolve returns a solution that contains a RootOf expression.

Consider the following ODE and initial condition.

The RootOf expression in the solution for this ODE tells us that an explicit solution for the ODE could not be found. (Note that odetest verifies that the solution does satisfy the ODE.)

There are two ways to solve this. The first is to call dsolve with the implicit option to force dsolve to return an implicit equation for y(x).

The second is to use the DEtools:-remove_RootOf command.

To graph the solution, use the plots:-implicitplot command.

For this example we can use eval  and fsolve to find the value for x given a value of y(x).

This example shows how to use dsolve get a numerical solution to $\frac{\ⅆ}{\ⅆ x} y\= -y$.

Define a variable called basicOde, a set that contains the ODE and the initial condition, y(0)=5.

To get a numerical solution, use dsolve with the numeric parameter.

The output is a procedure that you run to analyze the solution to your ODE. In the output expression, the x_rkf45 term tells you the numerical method used in the solution (in this case rkf45, a Fehlberg fourth-fifth order Runge-Kutta).

To get a value for your solution at a point, run the procedure with the point as the parameter. The following command finds the value for y when x=5.0. The output is a list of equations. The first element in the list echoes your input. The second element gives you the value of y(x).

Use the plots:-odeplot command to graph your solution. The odeplot command accepts the solution from dsolve/numeric as its first parameter. The following commands plot sol over x=0..5.

Note that this solution is very specific in terms of the initial conditions and the terms in the ODE. The next section shows how you can generalize the solution by using the parameters option with dsolve/numeric.

Assigning the Solution to a Variable

The procedure that dsolve returns is useful for finding the value of the solution at a point and plotting the solution with odeplot, but it cannot be used in fsolve. For example, let's say you want to find the value of x for which  y(x)=2. Using sol directly in fsolve returns the same expression unevaluated.

To solve this, use the output=listprocedure option with dsolve, and then assign the solution to a variable with the eval(expression,equations) command.

The following commands verify that the output from sol and Y are identical (and that the result from fsolve is correct).

With the solution assigned to a variable, you can use the standard plot command to graph your solution instead of plots:-odeplot.

This example shows how to use dsolve get a numerical solution to $\frac{\ⅆ}{\ⅆ x} y\=a\cdot y$ with the initial condition y(x0) = y0. The terms a, x0, and y0 are parameters that you will be able to set after the solution has been generated.

Define a variable called basicOde2, a set that contains the ODE and the initial condition, y(x0) = y0.

Use dsolve to get a numerical solution. Also, because basicOde2 contains the parameters a, x0, and y0,use the parameters option.

Before you can work with the solution, you must initialize the parameters a, x0, and y0. To initialize the parameters, run the output procedure with values inserted in the parameters list. The following sets a to -1, x0 to 0, and y0 to 5.

To get a value for your solution at a point, run the procedure with the point given as the argument. The following command finds the value for y when x=5.0. The output is a list of equations. The first element in the list echoes your input. The second element gives you the value of y(x).

The following graphs sol2 over x=0..5.

At any time you can change the values for the parameters to get a different solution for your system.

Finally, if you want to analyze the solution using fsolve, you must generate a solution using the output=listprocedure option, and then assign the solution to a variable.

Before you can use the function, you must set values for the parameters a, x0, and y0.

You can now use fsolve on the solution. The following finds the value of x for which the solution equals 2.

Define the system of ODEs along with their initial conditions. The system of ODEs in this example is the Lotka-Volterra predator-prey system, where x(t) represents the population of the prey and y(t) the population of the predator.

Generate the solution with a list of parameters.

Before working with the solution, set values for the parameters.

To find the populations at a particular time, run the solution with the time given as the argument. The result is a list of equations for the independent variable (t), the prey population (x(t)), and the predator population (y(t)).

To graph multiple curves with odeplot, use a nested list that contains a list for each pair of coordinates you would like to graph. For example, for the following odeplot command graph, both the prey and predator populations are displayed on the same graph.