A Formula for Some Particular Solutions

I found this a few days ago, and while I may call it obnoxious, I wouldn’t necessarily call it useless (or at least as useless as I thought it was when I originally derived it).

It is a general form for a particular solution of differential equations of the forms

The more compact way to write all of these is

Where \(D\) is the differential operator \(Dy=y'\), \(p(x)\) is the characteristic equation of the differential equation, and \(\alpha=a+bi\) for the cases with sines and cosines.

The formula is as follows:

Let \(s\) be the smallest integer such that \(p^{(s)}(\alpha)\neq0\).

A particular solution is given by

For the case \eqref{re} take the real part of the above expression, and for \eqref{im} take the imaginary part.

Now I know what you’re thinking: “Oh dear god what is that hidious and disgusting thing”

When would this monstrosity save time?

Here’s the deal: This formula is only going to save you time if \(m\) is small and \(n\) is large. That is to say when the order of the equation is high and the highest power on the forcing function is low.

The reason for this is relatively straightforward: This formula relies on computing a determinant, and big determinants suck to evaluate. If \(m\) is small, then the determinant is small. If \(m=1\), it’s just a \(3\times 3\) matrix which can be evaluated relatively easily.

Take an example:

Undetermined coefficients would be the most efficient method for solving this, but would require you to:

1. Solve the homogeneous equation (which involves factoring a 7th degree polynomial)
2. Differentiate a function of the form \(t^s(A+Bt)e^t\) seven times
3. Plug in that mess of terms and simplify
4. Solve a system of equations

The system of equations would be in two variables which isn’t that bad but I don’t want to differentiate \(t^s(A+Bt)e^t\) seven times and simplify that mess.

In contrast, to evaluate the formula you would simply

1. Evaluate \(p(1),p'(1),\ldots\) in total \(s+m\) times (for this specific problem, \(s=2\) so you’d just need to go up to \(p'''(1)\)). Plus, this nets you \(s\) homogeneous solutions for your trouble.
2. Evaluate \(\frac{(2+1)^1(2+2)^{1-1}}{1^12^{1-1}}\)
3. Evaluate 3 binomial coefficients (only one of which is not just \(1\))
4. Evaluate a \(3\times 3\) determinant (with a lot of zeros)

Note that if \(m=0\), however, then this does just reduce down to the Alpha Method, which is far more straightforward and does always save time.

And finally, you can use this for forcing functions like \((b_0+\ldots+b_mt^m)e^{at}\sin(bt)\) and \((b_0+\ldots+b_mt^m)e^{at}\cos(bt)\). All you need to do is use \(\alpha=a+bi\), and then take the real part if it was originally cosine and the imaginary part if it was sin.

An example

Let us actually go through the process of applying this formula to

Here we identify \(\alpha=1\) and \(m=1\). We see that \(b_0=0\) and \(b_1=1\). Additionally,

Our particular solution will then be, for some \(s\) yet to be determined,

We can easily compute \(\frac{(s+1)^1(s+2)^{1-1}}{1^m2^{1-1}}=s+1\), but most importantly we can already expand along the third column. This eliminates any need to compute binomial coefficients and reduces the determinant to be \(2\times 2\). Then our particular soultion will be

A quick remark: The fastest way to evaluate polynomials is actually Synthetic Division. In fact, it would be the fastest way to evaluate multiple derivatives at the same point (by repeatedly doing synthetic division). I hope to someday make a post on that method in the future. (EDIT: I did)

Regardless, one can calculate that \(p(1)=0\). This actually tells us a homogeneous solution is \(y_1=e^t\). So now if we go back to solving the homogeneous equation it’s only a 6th degree polynomial.

Take the derivative:

We find that \(p'(1)=0\) as well. Well at least we get another homogeneous solution \(y_2=te^t\).

Another derivative:

\(p''(1)=16\) and at last we find our \(s=2\), and by extension the most important part of our solution \(p^{(s)}(\alpha)=16\). All that is missing is \(p^{(s+1)}(\alpha)\), for which we need \(p'''(1)\):

And, finally, \(p'''(1)=120\). We now have everything we need:

And this is indeed the correct answer!

What is this sorcery, magic man?

The mechanics behind this formula involve the technique of Function Interpolation, that I’ve previously written about, and the exponential shift identity’s interaction with Taylor series.

The exponential shift identity is as follows

For intuition on why this works, observe that

You can prove using induction that \(D^n(e^{\alpha t}f(t))=e^{\alpha t}(D+\alpha)^nf(t)\). \eqref{eshift} then follows because differentiation is linear.

We know that our particular solution to \eqref{1} will be of the form

where \(s\) is the multiplicity of the root \(\alpha\) in \(p(x)\) so that we do not duplicate any homogeneous solutions (Note that this implies \(p^{(k)}(\alpha)=0\) for \(0\leq k\leq s\)). Keep in mind that \(s\) could be zero.

We will use the shorthand \(c(t)=c_0+c_1t+\ldots+c_mt^m\) and \(b(t)=b_0+b_1t+\ldots+b_mt^m\).

Our exponential shift formula (vaguely) tells us what will happen when we plug in our particular solution to the differential equation:

Note that normally I would write \(y_p\) in sigma notation, but I found that it just makes things more complicated to do so.

Using Taylor Series, we can write \(p(x)\) as

Note: We begin at \(k=s\) because we assume that \(p^{(k)}(\alpha)=0\) for \(0\leq k\leq s\).

Why would we do this? Well, because

Therefore,

Now we can simply deal with derivatives and values of \(p^{(k)}(\alpha)\).

Now, since \(t^sc(t)\) is of degree \(s+m\), any power of \(D\) greater than \(s+m\) will disappear. Therefore, we can change \(\frac{p^{(s)}(\alpha)D^s}{s!}+\ldots+\frac{p^{(n)}(\alpha)D^{n}}{n!}\) to

Observe that this is still true if \(m>n\), since this, in that case, is equivalent to just adding a bunch of zeros to the sum.

Now our goal is to have the output give us \(b_0+b_1t+\ldots+b_mt^m\), so we should group up the terms of each power.

The constant term, we will have

Consider that \(D^kt^j=\begin{cases}\frac{j!}{(j-k)!}t^{j-k},&k\leq j\\0,&k>j\end{cases}\).

Now you may be wondering why I’m bothering to write the \(0!\) and not cancelling the factorials. The reason is that we happen to have our good friends: the binomial coefficients!

\begin{equation} \binom{n}{k}=\frac{n!}{k!(n-k)!} \end{equation}

So we can write this instead as

And in general, to find \(b_k\) (\(0\leq k\leq m\)),

We start with higher and higher coefficients \(c_j\) because when taking \(s\) or more derivatives, only terms with power \(s+k\) (corresponding to \(c_k\)) or higher will yield terms with power \(t^k\).

So we may write the system of equations for our coefficients of \(c(t)\) in matrix form as

So here is where the determinant polynomial comes in. We have a system of equations corresponding to the coefficients of a linear function (a polynomial). To save a lot of space, we will write the above equation as

We can write \(c(t)\) using the determinant polynomial as

The good news is that the coefficient matrix is upper triangular, so its derivative is actually relatively simple. The determinant of the matrix is in fact

You can use the definition of the binomial coefficients to show that this is equal to

And… that’s it. Substituting the results into \eqref{pform} yields \eqref{2}.