Two entirely different topics here:
- How do we write out systems of differential equations and solve
them?
Let's step through an example slowly. We'll rewrite the
differential equation
x'' + 2x' + x = sin(t)
in the form of a system. To do this, we introduce another
variable, which I will with infinite imagination call y:
let y = x'. Then, substituting y
everywhere we see an x' in the original equation, we
obtain the system of equations
x' = y
y' = -x - 2y + sin(t)
The book considers the systems in terms of the subscripted
dependent variables x1,
x2, etc. In this case,
we'd be taking x1 = x and
x2 = y, to
get the system
x1' =
x2
x2' =
-x1 - 2
x2 + sin(t)
It doesn't matter which of these two notations you choose, so
long as you're consistent and can see how they are equivalent.
Then, how are these written as vectors? Let's think about
grouping the left- and right- hand sides of the equations in the
system more tightly:
| [ |
x1' |
] |
= |
[ |
x2 |
]
|
| x2' |
-x1 - 2
x2 + sin(t) |
The entries in the [ ] (square brackets) are the
column vectors. So the column vector x' is the vector on
the left-hand side, and that on the right-hand side is the
column vector f(t, x1, x2).
Great! Now, what about solving these? For this problem we know
how to solve it because we can solve the original second-order
differential equation. The solution to this is xg =
xu +
xd, or
xg =
(C1 +
C2 t)
e-t - (1/2) cos(t)
We said that x was x1 and x' was x2, so, taking the derivative of
this, we have the solution vector
| [ |
x1 |
] |
= |
[ |
(C1 +
C2 t)
e-t - (1/2)
cos(t)
|
]
|
| x2 |
(-C1 +
-C2 t +
C2)
e-t + (1/2)
sin(t)
|
- How do we find the solution to a problem with sines and cosines
in the driving term?
For example, the previous problem (a coincidence?). Let's solve
x'' + 2x' + x = sin(t)
For the undriven problem, x'' + 2x' + x =
0, notice that this is the same as
P(D)[x] = (D + 1)2[x] = 0
Plugging in x = er
t, we get
P(D)[er t] = (D
+ 1)2[er
t] = 0
Then we know that P(D)[er
t] = er
t P(r), so
er t (r +
1)2 = 0,
or r = -1 (twice; it's a repeated root). So the
undriven solution is
xu =
(C1 +
C2 t)
e-t
Then, to find the driven solution we can proceed in one of two
ways:
- Notice that the driving term f(t) =
sin(t) is the same as Imag(ei t). Thus, if we find a
solution to
z'' + 2z' + z = ei t,
the solution we want, xd, is just
Imag(z).
Let's do this. We want z to be an exponential, so
guess z = A ei
t. Plugging in, we get
P(D)[z] = (D + 1)2[z] = ei t,
or
(D + 1)2[A ei t] = ei t, so
A ei t (i +
1)2 = ei t.
Dividing by ei t
and solving for A, we get A =
1 / 2i. (Because (i+1)2 = i2 + 2i + 1 =
-1 + 2i + 1.) It's convenient to
multiply the numerator and denominator of A by
i to get A = -i / 2.
So
z = -(i / 2) ei t = -(i /
2)(cos(t) + i sin(t)) =
(1/2)(-i cos(t) + sin(t))
We want the imaginary part of this, which is the cosine.
Thus
xd = -(1/2) cos(t)
- Option two is to just guess
xd = A
cos(t) + B sin(t)
plug it in to the equation, and find what A and
B work. Notice that we need to start with both the
sine and the cosine term in most cases.
The completion of this method is left as an exercise for
the reader...