Hw7

Week 7 Problems: 1)

Find the general solution in real form to  − 3 0 2   x′ =  1 − 1 0  x . − 2 −1 0   Solution:

First find the eigenvalues: −3−λ 1 −2

0 −1 − λ −1

2 0 = −λ3 − 4λ2 − 7λ − 6 = 0 ⇒ λ = −2, − 1 ± i 2 −λ

Next find the eigenvectors:

λ = −2 :  − 1 0 2  x1   0   2         1 0  x 2  =  0  ⇒ x1 = 2 x3 , x1 = − x 2 ⇒ x =  − 2   1  − 2 − 1 2  x   0   1    3      is an eigenvector. λ = −1 + i 2 : − 2 − i 2   1   −2

0 2  x1   0       2  x1 , x1 = i 2 x 2 −i 2 0  x 2  =  0  ⇒ x3 = 1 + i  2       − 1 1 − i 2  x3   0   i 2   0   2        =  1  + i 0  ⇒ x= 1        − 1 + i 2   − 1  2  The general solution is:  0    0    2  2  2              − 2t −t    x = C1  − 2 e + C 2  1  cos 2t −  0  sin 2t e + C 3  1  sin 2t +  0  cos 2t e −t            1    − 1    − 1  2 2          

( )

( )

2)

( )

Find the solution to the initial value problem

( )

( )

 1 0 0  −1      x ′ =  − 4 1 0  x, x ( 0 ) =  2   3 6 2  − 30      Solution: Eigenvalues are λ=1, 1, 2 by inspection. The eigenvectors are:

λ = 2:  − 1 0 0  x1   0   0         − 4 − 1 0  x 2  =  0  ⇒ x1 = x 2 = 0, x3 arbitrary ⇒ x =  0   3 1 6 0  x3   0     is an eigenvector.

λ = 1:  0 0 0  x1   0   0          − 4 0 0  x 2  =  0  ⇒ x1 = 0, 6 x 2 + x3 = 0 ⇒ x =  1   3 6 1  x   0   − 6   3      is an eigenvector and there are no other linearly independent eigenvectors. Two solutions are: 0  0    2t   ( 2) x =  0 e , x =  1 e t 1  − 6     Look for another solution of the form ( 1)

 0 0 0  u1   0       1 x = t ⋅ x + ue ⇒  − 4 0 0  u 2  =  1  ⇒ u1 = − , 3u1 + 6u 2 + u 3 = −6 4  3 6 1  u   − 6    3     1  1 −  −  0      4  4 ⇒ u =  0  ⇒ x ( 3) =  1 te t +  0 e t  − 6  − 21   − 21     4  4     is another solution. Therefore ( 3)

( 2)

t

  1   0  −   0  0    2t   t   t  4  t  x = C1  0 e + C 2  1 e + C 3   1 te +  0 e  1  − 6  − 6  − 21        4        is the general solution. Applying the initial conditions gives 1   − C3    −1  4     C2   =  2  ⇒ C 3 = 4, C 2 = 2, C1 = 3  21     C1 − 6C 2 − C 3   − 30  4     1   0  −   0  0    2t   t   t  4  t  ⇒ x = 3 0 e + 2 1 e + 4  1 te +  0 e  1  − 6  − 6  − 21          4       is the solution. 3)

Let Φ be a fundamental matrix for the problem y ′ = Ay that has the property that   0   I . Show that Z1    t    s  and Z 2    t  s  both satisfy the initial value problem Z   AZ , Z  0, s     s  .

Solution: Since   t  is a fundamental matrix for y   Ay we have   t   A  t     t    s   A  t    s      t    s     A  t    s  . Therefore   t    s  satisfies the equation. We also have   0   s   I   s     s 

For the initial condition. Since   t  is a fundamental matrix for y   Ay we have d  t  s  A  t  s  . dt Therefore   t  s  satisfies the equation. We also have

 ts

d    A    . Let d

  t  s

t 0

   s

so exactly the same initial condition is satisfied. Since linear initial value problems have unique solutions and we find that   t  s  and   t    s  satisfy the same initial value problem we can conclude that they are equal. 4)

Find the general solution to  e −2t   1 1    y   y    t   4  2   2e  by the method of variation of parameters as described in our text and by the method of variation of parameters using the special fundamental matrix Φ as described in class. Solution: A fundamental matrix for the homogeneous equations is easily found and is  e −3t Y =  − 3t  − 4e

e 2t   1 1   X =  2 t  with a corresponding matrix of eigenvectors e   − 4 1  1  − 3t 1 2t . General solutions are found in the form y = C1  e + C 2  e + y p  − 4 1 where now we find the particular solutions by various methods. “Book’s Method”:  1 3t  e = 5  4 e −2t  5

1   1 t 2 4t  − e 3t  − 2t  e + e    e 5  ⇒ Y −1  5   5  Y −1 =  − 2e t   4 − 4t 2 −t  1 −2t    e   e − e  5 5   5  1 t 1 4t   1 t  − 2t − 2t  e + e     e  −1  e −1  e 5 10       ⇒ ∫Y  dt = ⇒ yp = Y ∫Y  dt =  2  t  t   − e −2t   − 1 e − 4t + 2 e −t   − 2e   − 2e      5   5 “Φ”- Method: 1  Y −1 ( 0 ) =  5 4  5

1  1 − 3t 4 2 t −   e + e 5  ⇒ Φ = Y ( t )Y −1 ( 0 ) =  5 5 1  4 − 3 t  − e + 4 e 2t   5  5  5

1 1  − e −3t + e 2t  5 5  4 − 3t 1 2 t  e + e  5 5 

 1 −3( t − s ) 4 2 ( t − s ) + e  e 5 yp = ∫ 5 4 4 − 3 ( t − s ) 0 − e + e 2( t −s )  5  5  1 t 3 −3t 1 2t   e − e − e  10 5  = 2  − e − 2 t + 6 e − 3t − 1 e 2 t    5 5   t

1 1  − e −3( t − s ) + e 2 ( t − s )  e − 2 s  5 5  ds 4 −3( t − s ) 1 2( t − s )  − 2e s  e + e  5 5 

(Note: the last particular solution varies from the other by the addition of homogeneous solutions and is still a valid particular solution.) 5)

Use the method of diagonalization to find a particular solution to Problem 4. Solution: We can use the eigenvalues and eigenvectors found above. 1 1  1  1  1 2t 2 t      2t e  e     1 1 5 5 5 5  e   5 5 1 1 X      X   4 1   X g   4 1  t   4 1 4     2e   e 2t  2 et          5   5 5   5 5   5  1 2t 2 t  e  e   u u  3 0  1    1  5 5          0 2   u2   4 e 2t  2 et   u2    5   5 Let 

1 1  2 A  3 A  5  A  5

u1  Ae 2t  Bet  

 B  3 B  2  B  1  5 10 4 1  2C  2C   C     5 5 u2  Ce 2t  Det    D  2D  2  D  2  5 5  1 2t 1 t  e  e   1 t  e  1 1  5 10 y p  Xu     2    4 1   1 e 2t  2 et   e 2t      5   5