Solving Non-homogeneous ODE's: Method of Undetermined Coefficients ...

In my last post, I stated something rather paradoxical about solving non-homogeneous differential equations: ...to find a solution to the non-homogeneous equation, we need to find a solution to the non-homogeneous equation.

A bit of a 'Catch-22'? Well, in this post we'll walk through the first of such methods...

Figure 1. Graph of the particular solution to equation (6)

The method of undetermined coefficients is a simple, but effective way of finding solutions to the non-homogeneous equation. Here, we use more or less a trial-and-error approach to find solutions for .

Rather than explaining how it's done, first let's just go through a few examples to understand the mechanics of finding these solutions. After that, we'll summarise the method. So let's go straight to Example 1...

Example 1

Let's find the particular solution to the equation (I started with the number (6), as it follows equation (5) from the last post)...

With initial conditions...

First, we must find the solution to the homogeneous equation...

...the characteristic equation of which is...

...giving rise to the roots...

...and thus the general solution to the homogeneous equation is...

Ok, so now that we have a solution to the homogeneous equation, we need to find a solution to the non-homogeneous equation.

How do we go about it?

Well, let's try a solution of the form that is similar to the term on the right-hand side of equation (6). Let's try...

...thus and . Let see how this goes when we substitute these expressions into (6)...

Now, equating the coefficients, we see that 3C = 18 (i.e. C = 6), but 2C = 0 as well (i.e. C = 0). This is a conflicting, or contradicting result.

So that obviously didn't work, but let's not give up here.

Let's try including some of the lower degrees to turn the trial solution into a polynomial...

Now, the first and second derivatives are...

..and...

And now substituting these into (6)...

So equating the coefficients again, we have...

...we also have...

...and finally, we have...

Therefore, our solution for is...

So, from (3) of the last post, the general solution is...

In similar fashion to previous methods of finding particular solutions, we apply the initial conditions. Firstly...

To apply the second initial condition, we need to find the first derivative...

And therefore, the final particular solution is...

Figure 1 above shows a graph of this solution. The solution is simply a superposition (i.e. a fancy way of saying 'addition') of an oscillating (circular) component: ; and a polynomial component .

As you can see by the red curve, the solution is dominated by the polynomial.

We will go through another example in the next post.

Credits:

All equations in this tutorial were created with QuickLatex

First Order Differential Equations

First order Non-linear Differential Equations

Second Order Differential Equations

Introduction to Second Order Differential Equations
Finding a Basis for solutions of Second Order ODE's
Roots of Homogeneous Second Order ODE's and the Nature their Solutions
Modelling with Second Order ODE's: Undamped Free Oscillations
Modelling Car Suspension with ODE's: Damped Free Oscillations Part 1
Modelling Car Suspension with ODE's: Damped Free Oscillations Part 2
Modelling Car Suspension with ODE's: Damped Free Oscillations Part 3
Non-homogeneous Differential Equations
Solving Non-homogeneous ODE's: Method of Undetermined Coefficients Part 1

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Solving Non-homogeneous ODE's: Method of Undetermined Coefficients Part 1

Example 1

First Order Differential Equations

First order Non-linear Differential Equations

Second Order Differential Equations