8.7 Systems of firstorder linear differential equations
Two equations in two variables
Consider the system of linear differential equations (with constant coefficients)x'(t)  = ax(t) + by(t) 
y'(t)  = cx(t) + dy(t). 
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(t) and y'(t) 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 secondorder 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 secondorder equation in y(t).)
 Example

Consider the system of equations
x'(t) = 2x(t) + y(t) y'(t) = −4x(t) − 3y(t). x"(t) − 2x'(t) = −4x(t) − 3x'(t) + 6x(t),orx"(t) + x'(t) − 2x(t) = 0.We have seen that the general solution of this equation isx(t) = Ae^{t} + Be^{−2t}.Using the expression y(t) = x'(t) − 2x(t) we gety(t) = Ae^{t} − 2Be^{−2t} − 2Ae^{t} − 2Be^{−2t},ory(t) = −Ae^{t} − 4Be^{−2t}.
General linear systems
We may write the systemx'(t)  = ax(t) + by(t) 
y'(t)  = cx(t) + dy(t) 

= 


. 
x'(t) = Ax(t),
where A is an n × n matrix.
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) = Ce^{At}.
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 isYou will immediately ask: what is e^{At} when A is a matrix? Recall that if a is a number we havex(t) = Ce^{At}.
e^{a} = 1 + a/1! + a^{2}/2! + ...,
or, more precisely,
e^{a} = ∑∞
k=0(a^{k}/k!).
Now, when A is a matrix we may define
k=0(a^{k}/k!).
e^{A} = ∑∞
k=0(A^{k}/k!),
where A^{0} 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 e^{At}. Therein
lies the difficulty. Techniques exist for finding e^{At}, but they involve methods more advanced than the ones in this tutorial.
k=0(A^{k}/k!),
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 secondorder equation.
 Example

Consider the system
x'_{1}(t) x'_{2}(t) = 0 −b b 0 x_{1}(t) x_{2}(t) . A = 0 −b b 0 . A^{k} = 0 (−1)^{(k+1)/2}b^{k} (−1)^{(k−1)/2}b^{k} 0 , A^{k} = (−1)^{k/2}b^{k} 0 0 (−1)^{k/2}b^{k} . e^{A} = 1 − b^{2}/2! + b^{4}/4! − b^{6}/6! + ... −b + b^{3}/3! − b^{5}/5! + ... b − b^{3}/3! + b^{5}/5! − ... 1 − b^{2}/2! + b^{4}/4! + b^{6}/6! + ... . e^{A} = a_{11} a_{12} a_{21} a_{22} a_{11} = 1 − b^{2}/2! + b^{4}/4! − b^{6}/6! + ... a_{12} = −b + b^{3}/3! − b^{5}/5! + ... a_{21} = b − b^{3}/3! + b^{5}/5! − ... a_{22} = 1 − b^{2}/2! + b^{4}/4! + b^{6}/6! + .... sin b = b − b^{3}/3! + b^{5}/5! − ...andcos b = 1 − b^{2}/2! + b^{4}/4! − b^{6}/6! + ...Thuse^{A} = cos b −sin b sin b cos b . x_{1}(t) = C_{1}cos bt − C_{2}sin bt x_{2}(t) = C_{1}sin bt + C_{2}cos bt.
e^{ix} = cos x + isin x
(where i 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 (1974).