Homogeneous Differential Equations

Solving Homogeneous Equations

Homogeneous differential equations are the fundamental building block for all other differential equations you will be solving in APMA 33. You will recall on the previous page that we can easily write a characteristic equation to model a differential equation. In this section will will learn how to solve the most elementary types of these characteristic equations. In general, a homogeneous equation is one where the left-hand side of the equation is set equal to 0 on the right hand side of the equation. For instance, $x^2-4=0$ is a homogeneous equation, and similarly $8\frac{dY}{dt} + 14y = 0$ is a homogeneous differential equation. The specific type of homogeneous differential equation is the ordinary linear differential homogeneous equation, which will always yield a polynomial characteristic equation.

The easiest way to solve a differential equation of this type is to follow the following system:

1. Write the characterisitic equation of the differential equation
2. Solve the characterisitic equation for its roots
3. Analyze the multiplicity and Real/Complex value of the root
4. Write the general solution in terms of arbitrary constants

Characteristic Equation

Going one step further, we can simply talk about a general class of differential equations with constant coefficients by looking at their characteristic equations. Much like the big \( \texttt{D} \) operator, we will desbribe abstract differential equations with what is known as the characteristic polynomial. We will use either the Greek letter \( \lambda \) or the Latin letter "r" to write a polynomial equation that relates to a differential equation. Therefore, solving homogeneous constant coefficient equations is completely reduced to an algebraic equation. Being able to write the characteristic equation of a linear differential system will make it much easier to solve many types of differential equations. The characteristic equation is simply a set of differential operators that are abstracted from the differential equation you are given. Below is a quick derivation of how to associate the differential equation with an operator equation and then with the polynomial.

Given the differential operator applied to a function \( y(t) \) :

We can rewrite it as which we associate with the polynomial And similarly
\( 3 \frac{{\text d}^3 X}{{\text d}t^3} -4 \frac{{\text d}X}{{\text d}t} +9X \)
becomes
\( \left( 3\texttt{D}^3 -4\texttt{D} + 9\right) X .\)
The corresponding characteristic polynomial becomes
\( 3\lambda^3 -4\lambda + 9 .\)

Soon enough, you will be able to look at any differential equation and write its charactistic equation very quickly. Since we can easily write polynomial equations in MuPAD, writing the characteristic equation is easy - it as simple as writing a normal polynomial equation.

Analysis of Characteristic Equation Roots

You should already know how to write the characteristic equation from a differential equation, which you learned on the previous page. Once you've written that equation, it should be a polynomial function that is set equal to zero. Now you can solve for the roots of the characteristic equation by hand or using MuPAD. These roots are characterized with the following properties:

• Multiplicity: How many copies of each the root are there?

  Example: \( r^6-7r^5+9r^4+27r^3-54r^2 \) reduces to \( r^2(r+2)(r-3)^3 .\) The multiplicity of root r=0 is two, r=-2 is one, and r=3 is three.

• Real or Complex Value Is r real valued or complex?

  Example: \( \lambda^3+4\lambda =0 \) yields the roots \( \lambda =0,\ \lambda =2{\bf j}, \) and \( \lambda =-2{\bf j} , \) <\span> where \( {\bf j}^2 = -1 . \) Here \( {\bf j} \) is the unit vector on the complex plane in the positive vertical direction.

Writing the Solution

Solutions for these kinds of linear ordinary differential equations will come in the form of \( C\,t^{m-1}\,e^{\lambda t} \) for real valued roots with multiplicity m and \( t^{m-1}e^{t\,\Re\lambda} \left( C_{1}\,\cos(\mu t)+C_{2}\,\sin{\mu t} \right) \) for complex roots of multiplicity m of the form \( \lambda = \Re \lambda + \mu{\bf j} , \) where \( \mu = \Im \lambda \) is the imaginary part of \( \lambda \) Here e is Euler's number (approximately 2.71828). A great proof for why these equations are the true solutions can be found here. The C in both general solutions are arbitrary constants. We need these because we are abstractly talking about a system of solutions for the differential equation. You will learn more about this in a few pages on this tutorial. We will multiply by \( t^{m -1} , \) where multiplicity is denoted as m because we want each root's solution to be linearly independent from each other. Multiplying by another t will always guarantee a new linearly independent solution. If you are unsure, you can try to calculate the wronskian of two solutions to check.

Using MuPAD to Solve

The following examples will illustrate how you might accomplish solving some of these homogeneous equations using MuPAD to help you with the more difficult computation. We will be using the "solve()" function as well as the other tools you have learned so far to do this.

Example 1: Solve 3y' -y =0.

 

We know that the characteristic equation is

We know that the only root, 1/3 is real.  To check the multiplicity of the characteristic equation we can factor it

Thus the multiplicity of our root is 1.

We can now write the answer in the form:

Where exp() tells MuPAD to make an exponent of Euler's number e

 

 

Example 2: Solve y'' - 6y + 2y =0

 

Where Full is tells MuPAD to accept non-integer factors

 

The final answer:

 

Example 3: Solve y'' -9y' +12y =0

First, write the characteristic equation: or, using the Greek letter \( \lambda \)

Then write the roots of the characteristic equation and factor to check the multiplicity of the characteristic equation:

With next command, we factor the characteristic polynomial. Note: 'Full' allows for non-integer factors to be accepted by MuPAD.

Final solution:

 

Example 4: Solve y'''+4y'=0

Final Answer:

We can factor out the real and imaginary parts of the roots using Re for real and Im for imaginary extraction

We know that e^0 =1, so we end up with

 

Example 5: solve 2y''''+11y'''+18y''+4y'-8y=0

Notice that the root r=2 has multiplicity 3, so we must include the root 3 times and each time they must be linearly independent from each other, so we will multiply the 2nd result by t

and the third result by t^2

 

Final Answer:

 

Example 6: solve y''''+16y =0

Final Answer

Root 1

Root 2

Root 3

Root 4