First Order Linear Differential Equations16
3.7 Nonhomogeneous Systems of Differential Equations
The general systems of differential equations with constant coefficients can be written as
Pi1(D) x1 + Pi2(D) x2 + . . . + Pin(D) xn = fi(t), 1 ( i ( n (3.7-1)
The above system can be written in the compact matrix form P(D)x = f with the following definitions of matrix operator P(D), vector x, and vector f
P(D) = [pic], x = [pic], and f = [pic]
The following is an example of a general system of differential equations with constant coefficients.
[pic] + 2x + [pic] + 6y = 2et
2[pic] + 3x + 3[pic] + 8y = (1
The system can be rearranged to
[pic]x + [pic]y = 2et
[pic]x + [pic]y = (1
In terms of the differential operator
(D + 2)x + (D + 6)y = 2et ( P11(D)x + P12(D)y = 2et
(2D + 3)x + (3D + 8)y = (1 ( P21(D)x + P22(D)y= (1
Consider the system (3.7-1) in matrix form
P(D)x = f(t) (3.7-1)
The associated homogeneous system of equations is
P(D)x = 0 (3.7-2)
The complete solution of (3.7-2) is called a complementary function of (3.7-1) and any particular solution of (3.7-1) is called a particular integral of that system. To find a complementary function, we assume solutions of the homogeneous system exist in the form
x = ke(t
Since Dr(e(t) = (re(t, we obtain
P(()e(tk = 0
or
P(()k = 0 (3.7-3)
Eq. (3.7-3) will have a nontrivial solution if
|P(()| = 0 (3.7-4)
This equation is called the characteristic equation of both the algebraic system (3.7-4) and the original differential system (3.7-2). For each root (i of the characteristic equation (3.7-4) there will be a solution vector ki of (3.7-3) determined to within an arbitrary scalar vector ci. For N distinct real roots (1, (2, . . ., (N.
x = c1k1[pic] + c2k2[pic] + . . . + cNkN[pic]
Example 3.7-1. Find a complete solution of the following first order system.
(D + 1)x + (D + 2)y + (D + 3)z = (e(t
(D + 2)x + (D + 3)y + (2D + 3)z = e(t (3.7-5)
(4D + 6)x + (5D + 4)y + (20D ( 12)z = 7e(t
Solution
The solution [pic]= [pic]e(t of the homogeneous system P(D)x = 0 exists if and only if
(( + 1)a + (( + 2)b + (( + 3)c = 0
(( + 2)a + (( + 3)b + (2( + 3)c = 0
(4( + 6)a + (5( + 4)b + (20( ( 12)c = 0
The characteristic equation of this system is
[pic] = (( ( 1) (( ( 2) (( ( 3) = 0
For ( = 1
2a + 3b + 4c = 0
3a + 4b + 5c = 0 ( [pic] = d1[pic] = c1[pic]
10a + 9b + 8c = 0
For ( = 2
3a + 4b + 5c = 0
4a + 5b + 7c = 0 ( [pic] = d2[pic] = c2[pic]
14a + 14b + 28c = 0
For ( = 3
4a + 5b + 6c = 0
5a + 6b + 9c = 0 ( [pic] = d3[pic] = c3[pic]
18a + 19b + 48c = 0
A complementary solution of (3.7-5) is therefore
[pic]= c1[pic]et + c2[pic]e2t + c3[pic]e3t
We now need to find a particular integral of the differential system (3.7-5). We choose a trial solution
v = [pic]e-t
Substituting v into (3.7-5), collecting terms, and canceling a factor of e-t
4a + 5b + 6c = 0
5a + 6b + 9c = 0 ( [pic] = [pic]
5a + 6b + 9c = 0
Hence a particular integral of (3.7-5) is
[pic]= [pic]e-t
A complete solution of the original system is
x = c1[pic]et + c2[pic]e2t + c3[pic]e3t + [pic]e-t
The constants c1, c2, and c3 can be determined from the initial conditions x(0) = 1, y(0) = 1, and z(0) = 1.
[pic][pic] = [pic]( [pic] = [pic]
The solution can be verified with MATLAB.
>>[x1,x2,x3]=dsolve('Dx+x+Dy+2*y+Dz+3*z=-exp(-t),Dx+2*x+Dy+3*y+2*Dz+3*z=exp(-t), 4*Dx+6*x+5*Dy+4*y+20*Dz-12*z=7*exp(-t)','x(0)=1,y(0)=1,z(0)=1')
x1 = 11/2*exp(t)+24*exp(2*t)-63/2*exp(3*t)+3*exp(-t)
x2 = -8*exp(2*t)-11*exp(t)+21*exp(3*t)-exp(-t)
x3 = -8*exp(2*t)+11/2*exp(t)+7/2*exp(3*t)
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