For each of the following operators T on R³ H 00 -2 0 1 (a) 10 1 Solve x' Tx where T is the operator in (a) Introduce the new basis (1, 0, 0), (0, –√2, √2), (1, -2, -1), and new coordinates (y₁, y2, y3) related to the old by The general solution is Therefore In the new coordinates the differential y₁ Y/₂ y3 = √2 y₂. X1 = x2 = x3 = Yı Y2 Y3 x1 = X2 X3 = = Y₁ + Y3, -√2 Y₂2 √2 Y2 equation becomes Y, - -√2 Y31 - 2y3, Y3. Cet, A cos (√2 t) + B sin (√2 t), -B cos (√2 t) + A sin (√2 t). Cet B cos (√2 t) + A sin (√2 t), (2B - A√2 ) cos (√2 t) - (B√2 + 2A) sin(√2 t), (B + A√2 ) cos (√2 t) + (B√2 − A) sin (√2 t).
For each of the following operators T on R³ H 00 -2 0 1 (a) 10 1 Solve x' Tx where T is the operator in (a) Introduce the new basis (1, 0, 0), (0, –√2, √2), (1, -2, -1), and new coordinates (y₁, y2, y3) related to the old by The general solution is Therefore In the new coordinates the differential y₁ Y/₂ y3 = √2 y₂. X1 = x2 = x3 = Yı Y2 Y3 x1 = X2 X3 = = Y₁ + Y3, -√2 Y₂2 √2 Y2 equation becomes Y, - -√2 Y31 - 2y3, Y3. Cet, A cos (√2 t) + B sin (√2 t), -B cos (√2 t) + A sin (√2 t). Cet B cos (√2 t) + A sin (√2 t), (2B - A√2 ) cos (√2 t) - (B√2 + 2A) sin(√2 t), (B + A√2 ) cos (√2 t) + (B√2 − A) sin (√2 t).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The answer has been given, please prove the process
question about ：APPLICATION OF COMPLEX
Transcribed Image Text:For each of the following operators T on R³
(a)
1 0 1
00
-2
0 1
0
Solvex'Tx where T is the operator in (a)
Introduce the new basis (1, 0, 0), (0, -√2, √2), (1, -2, -1), and new coordinates
(y₁, y2, Ys) related to the old by
In the new coordinates the differential equation becomes
y₁ = y,
y₂ = -√2 y3,
y3 = √2 Y₂.
The general solution is
x₁ = y₁ + Y3,
X1
X2
x₂ = -√2 y₂ - 2y3,
√2 Y2
Y3.
X3
Therefore
Yı
C'et,
Y₂ = A cos (√2 t) + B sin (√2 t),
y3 = -B cos (√2 t) + A sin (√2 t).
Y3
x₁ = Cet - B cos (√2 t) + A sin (√2 t),
X1
X2 =
X3
(B√2 + 2A) sin (√2 t),
(2B — A√2 ) cos (√2 t)
(B + A√2 ) cos(√2 t) + (B√2 − A) sin (√2 t).
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