8.7 Systems of first-order linear differential equations
Two equations in two variables
Consider the system of linear differential equations (with constant coefficients)
||= ax(t) + by(t)
||= cx(t) + dy(t).
We can solve this system using the techniques from the previous page
, as follows. First isolate y
) in the first equation, to give
y(t) = x'(t)/b − ax(t)/b.
Now differentiate this equation, to give
y'(t) = x"(t)/b − ax'(t)/b.
Why is this step helpful? Because we can now substitute for y
) and y
) in the second of the two equations in our system to yield
x"(t)/b − ax'(t)/b = cx(t) + d[x'(t)/b − ax(t)/b],
which we can write as
x"(t) − (a + d)x'(t) + (ad − bc)x(t) = 0,
an equation we know how to solve
Having solved this linear second-order differential equation in x(t), we can go back to the expression for y(t) in terms of x'(t) and x(t) to obtain a solution for y(t). (We could alternatively have started by isolating x(t) in the second equation and creating a second-order equation in y(t).)
Consider the system of equations
Isolating y(t) in the first equation we have y(t) = x'(t) − 2x(t), so that y'(t) = x"(t) − 2x'(t). Substituting these expressions into the second equation we get
||= 2x(t) + y(t)
||= −4x(t) − 3y(t).
x"(t) − 2x'(t) = −4x(t) − 3x'(t) + 6x(t),
x"(t) + x'(t) − 2x(t) = 0.
We have seen that the general solution of this equation is
x(t) = Aet + Be−2t.
Using the expression y(t) = x'(t) − 2x(t) we get
y(t) = Aet − 2Be−2t − 2Aet − 2Be−2t,
y(t) = −Aet − 4Be−2t.
General linear systems
We may write the system
||= ax(t) + by(t)
||= cx(t) + dy(t)
studied above, as
More generally, we may write a system of n
simultaneous homogeneous linear equations in the n
) for i
= 1, ..., n
x'(t) = Ax(t),
is an n
Now, if n = 1, in which case A is simply a number, we know that for the initial condition x(0) = C the equation has a unique solution
x(t) = CeAt.
Here's a stunning result:
for every value of n, the unique solution of the homogeneous system of linear differential equations x'(t) = Ax(t) subject to the initial condition x(0) = C is
x(t) = CeAt.
You will immediately ask: what is eAt
is a matrix? Recall that if a
is a number we have
ea = 1 + a/1! + a2/2! + ...,
or, more precisely,
ea = ∑∞
Now, when A
is a matrix we may define
eA = ∑∞
is the identity matrix (and 0! = 1). (You know how to multiply matrices together, so you know how to compute the right hand side of this equation.) That's it! You can now find the solution of any homogeneous system of linear differential equations ... assuming that you can compute the infinite sum in the definition of eAt
lies the difficulty. Techniques exist for finding eAt
, but they involve methods more advanced than the ones in this tutorial.
I give only one example, which shows how the trigonometric functions may emerge in the solution of a system of two simultaneous linear equations, which, as we saw above, is equivalent to a second-order equation.
You may wonder whether the emergence of the trigonometric functions in this example is accidental. It is not. Exponentials, the trigonometric functions, and complex numbers (which arise as roots of the characteristic equation in the technique of the previous page
), are closely related. Specifically, for any value of x
eix = cos x + isin x
is the square root of −1). But that is another story. An excellent, but advanced, exposition of the theory of systems of linear differential equations is contained in Hirsch and Smale