Math 2214
Solution Homework 10
Problem 1:
Given the following system of first order linear differential equations with solutions:
 2et 
e4t 
2 1 1
 0 
2
 t
 4t 



/
t 
Y  1 1 2 Y , solutions y1   e  , y 2   e  , y3  e  and Y(0)   0 
 e t 
 e 4t 
1 2 1 
 et 
 1
 
 
Using Abel’s Theorem:
a) Test to see if y1 is actually a solution
b) Use the Wronskian to see if the given solutions are a fundamental set.
c) Give the general solution and solve for the unknown coefficients.
Solution:
 2e t 
 2e t 
2 1 1
 
 
a) Y /  1 1 2  Y , test the solution Y1   et  , find Y1/   et 
 et 
 et 
1 2 1 
 
 
2 1 1
?
/
substitute to get Y1   1 1 2  Y1 ,
1 2 1 
 2et   2 1 1   2et 
 t ? 
 t
 e    1 1 2   e  ,
 et  1 2 1   et 
 
 
 2  2 1 1  2 
?
by Abel's Theorem at t = 0 we have  1  1 1 2   1 , and we have
 1 1 2 1   1
so we have that Y1 is indeed a solution
 2et

b) W = det  et
 et

2 2
 1   1
   
 1  1
e 4t 
 2 0 1
t
4t 
e
e   (use Abel's TH at t = 0) = det  1 1 1  6  0
 1 1 1
et e4t 
So we have a fundamental set.
0
 2et 
e 4t 
 0 
 
 
c) Y  C1  et   C2  e t   C3 e 4t 
 et 
e 4t 
 e t 
 
 
set up the augmented matrix using the initial value
 2e t 
e 4t 
 2 0 1 2  1 0 0 5 / 6 
 0 
 1 1 1 0   0 1 0 1/ 2  so Y  (5 / 6)  et   (1/ 2)  e t   (1/ 3) e4t 
 
 

 



t
 et 
e 4t 
 1 1 1 1 0 0 1 1/ 3 


e


 
 
Problem 2:
Put the following higher order differential equation y ///  ty //  t 2 y /  sin(t ) y  1/ t into
the form Y /  P(t )Y  G(t ) .
Solution:
 y1/   0
1
 / 
0
 y2    0
/
 y3    sin(t ) t 2
 
0   y1   0 
1   y2    0 
t   y3  1/ t 
Problem 3: Solve the following system of equations by finding the eigenpair.
 y1/  y1  2 y2
 /
 y 2  2 y1  y2
1) Set up the matrix Y /  AY
2) Set up the determinate det  A   I   0 and solve for the eigenvalues
3) Find the corresponding eigenvector for each eigenvalue in part 2).
4) Give the general solution Y  C1Y1  C2Y2
 1
5) Find the particular solution associated with the initial value Y (0    .
2
Solution
 y1/  1 2   y1 
1)  /   
 
 y2   2 1   y2 
2 
1  
2
2) det 
    2  3  0, so eigenvalues 1  3, 2  1
2
1




3) For 1  3 and find E1
 x1   x2 
 2 2 0  1 1 0 
1
 2 2 0   0 0 0  , This will give the vector  x    x   x2 1

 


 2  2
1 e3t 
so Y1  e3t     3t 
1 e 
4) For 2  1 and find E 2
 x1    x2 
 2 2 0  1 1 0 
 1
 2 2 0   0 0 0  , This will give the vector  x    x   x2  1 

 

 
 2  2 
 1  e  t 
so Y2  e  t      t 
1 e 
 e 3t 
 e  t 
The general solution is Y  C1  3t   C2   t 
e 
e 
 1
5) to find C1  C2 set up the augmented matrix for Y(0) =  
2
1 1 1 1 1 1 1 1 1  1 0 1/ 2 
1 1 2   0 2 3   0 1 3 / 2   0 1 3 / 2  so C1  1/ 2, C 2  3 / 2

 
 
 
