2. Solutions to 2nd-order ODEs¶
For first-order ODEs we introduced the Laplace transform, which converts a function of time \(t\ge 0\) to a function of an abstract \(s\). There’s no need for us to repeat its definition here, as we are more interested in applying its properties than deriving them.
First, we generalize the derivative property to the second derivative, which can be done by applying the original formula twice.
We collect the transforms in our earlier table and add a few new ones in the following table.
Function |
Transform |
|---|---|
\(x(t)\), \(f(t)\) |
\(X(s)\), \(F(s)\) |
\(x'(t)\) |
\(sX(s)-x(0)\) |
\(x''(t)\) |
\(s^2X(s)-sx(0)-x'(0)\) |
1 |
\(\dfrac{1}{s}\) |
\(e^{at}\) |
\(\dfrac{1}{s-a}\) |
\(e^{at} f(t)\) |
\(F(s-a)\) |
\(\cos(\omega t)\) |
\(\dfrac{s}{s^2+\omega^2}\) |
\(\sin(\omega t)\) |
\(\dfrac{\omega}{s^2+\omega^2}\) |
\(t^n, \quad n=1,2,3,\dots\) |
\(\dfrac{n!}{s^{n+1}}\) |
\(t^n f(t), \quad n=1,2,3,\dots\) |
\((-1)^n F^{(n)}(s)\) |
\(H(t-T)\) |
\(\dfrac{e^{-sT}}{s}\) |
\(H(t-T)f(t-T)\) |
\(e^{-sT}F(s)\) (shift theorem) |
\(\delta(t-T)\) |
\(e^{-sT}\) |
Note that complex numbers are allowed for the exponentials. Thus, for example,
Also, by Euler’s formula and linearity,
Of course, they agree:
2.1. Solving 2nd-order ODEs¶
Here is an example to get us started.
Example
Find a particular solution of \(x''-x=e^{-4 t}\).
Solution
Transforming both sides of the ODE leads directly to
so that
Since the goal is to find any particular solution, we are clearly happiest choosing one with \(x(0)=x'(0)=0\). Then
The method of partial fractions says that we can express this as
for constants \(A\), \(B\), \(C\). The easiest way to find them is to clear the denominators, yielding
The hard way to proceed is to multiply out the left side, then equate coefficients of like powers of \(s\). The easy way is to drop in \(1\), \(-1\), and \(4\) for \(s\):
We put this all together and find
Each of these terms is easy to invert, giving an exponential in each case:
Let’s pause for a moment. If we apply the method of undetermined coefficients to \(x''-x=e^{-4 t}\), we get \(x_p(t)=e^{-4t}/15\). The solution in the example was
Which is right? Both of them. Each method gives a particular solution, but they don’t give the same particular solution. Notice that the homogeneous solution is
and the two methods get different values for \(c_1\) and \(c_2\). Frankly, the MUC is simpler for this problem.
2.2. Poles¶
The basic transform \(\lx[e^{at}] = 1/(s-a)\), including for complex values of \(a\), leads us to look for roots of the denominator in a transform.
If
for some positive integer \(k\), complex number \(z\) such that \(F(z)\neq 0\), then \(z\) is a pole of order \(k\) of \(X\).
In the general case we get
If we choose to find a particular solution with zero initial conditions, then
The characteristic values of the ODE are therefore poles of \(X_p\), and the forcing function contributes one or more of its own poles.
Example
Find a particular solution of \(x''+9x = 15\sin(2t)\).
Solution
Taking the transform and solving for \(X\) gives
This has conjugate poles at \(\pm 2i\) and \(\pm 3i\).
If we want to keep everything real, the PFD is
We clear denominators to get
Equating like powers leads to \(A+C=B+D=9A+4C=0\) and \(9B+4D=30\). These imply \(A=C=0\), \(B=6\), \(D=-6\). Hence
Inverting the transform gives
In the previous example, we had
The first term on the right side leads to a combination of \(\cos(2t)\) and \(\sin(2t)\), while the second leads to a combination of \(\cos(3t)\) and \(\sin(3t)\). Since \(\cos(3t)\) and \(\sin(3t)\) form a basis of the homogeneous solution, we can leave them out of the inverse transform and still get a particular solution, \(3\sin(2t)\). We still have to keep the second term around in order to set up correct equations for \(A,B,C,D\); we just don’t need to solve for \(C\) and \(D\) or include their corresponding terms in \(x_p\). This can save a lot of algebra, though it basically returns us to the method of undetermined coefficients.