Page 330 - Advanced engineering mathematics

310 CHAPTER 10 Systems of Linear Differential Equations

If we write
⎛ ⎞
a
⎝
E 2 = b ⎠
c
then we have the system of algebraic equations
⎛ ⎞⎛ ⎞ ⎛ ⎞
0 −1 −5 a −1
⎝ 25 −5 0 ⎠⎝ b ⎠ = −5 ⎠ .
⎝
0 1 5 c 1
This system has general solution
⎛ ⎞
−α
⎝ 1 − 5α ⎠
α
for α any real number. Choose α = 1 to get
⎛ ⎞
−1
.
⎝
E 2 = −4 ⎠
1
This gives us a second solution of the differential equation
2 (t) = E 1 te −2t + E 2 e −2t
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
−1 −1 −1 − t
2t −2t −2t
= −5 ⎠ te + −4 ⎠ e = −4 − 5t ⎠ e .
⎝
⎝
⎝
1 1 1 + t
We need one more solution, linearly independent from the first two. Try for a third solution of
the form
1
2 −2t
3 (t) = E 1 t e + E 2 te −2t + E 3 e −2t .
2
Substitute this into X = AX to get

2 −2t
E 1 [te −2t − t e ]+ E 2 [e −2t − 2te −2t ]+ E 3 [−2e −2t ]
1
2 −2t
= AE 1 t e + AE 2 te −2t + AE 3 e −2t .
2
Divide e −2t and use the fact that AE 1 =−2E 1 and
⎛ ⎞
1
AE 2 = ⎝ 3 ⎠
−1
to get
⎛ ⎞
1
2 2
E 1 t − E 1 t + E 2 − 2E 2 t − 2E 3 =−E 1 t + ⎝ 3 ⎠ t + AE 3 . (10.2)
−1
Now
⎛ ⎞ ⎛ ⎞
−1 − 2(−1) 1
t = t.
E 1 t − 2E 2 t = (E 1 − 2E 2 )t = −5 − 2(−4) ⎠ ⎝ 3 ⎠
⎝
1 − 2(1) −1
Therefore, three terms cancel in equation (10.2) and it reduces to
E 2 − 2E 1 = AE 3 .
Write this equation as
(A + 2I 3 )E 3 = E 2 .

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October 14, 2010 20:32 THM/NEIL Page-310 27410_10_ch10_p295-342