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1. (10 points) Using the definition of Laplace transform, find L{-6t + 11} XXXXXXXXXXpoints) Use the Laplace transform to solve the following initial value prob-lem. No credit will be given for any...

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1. (10 points) Using the definition of Laplace transform, find L{-6t + 11} XXXXXXXXXXpoints) Use the Laplace transform to solve the following initial value prob-lem. No credit will be given for any other method y/ yt — 6y = —6t + 11, y(0) = —1, '00) = 4
3. (20 points) For each of the following boundary value problems, either show the problem has no solution, find its unique solution, or find all of its infinitely many solutions. (a) yil + y = x + 1. y(0) = 1, y =- (b) yfi + y = x + 1, y(7r) = 0, y(-77) = 0 (c) yti + y = x + 1, y(7r) = 0, y(277) = 3r XXXXXXXXXXpoints) Sketch three periods of the function defined below and find its Fourier series.
f (x)
x 0
Are there any values of x for which the series you found does not converge to the value dictated by the definition of f found above? To what value does the series converge in this case?
(25 points) Sketch three periods of the function defined below and find its Fourier series.
f(x) f(x ± 2) f(x) --{ i_x oIs this function even, odd, or neither? Explain your answer
Answered Same Day Dec 21, 2021

Solution

Robert answered on Dec 21 2021
128 Votes
1. Using the definition of the Laplace transform, find  6 11 .t L

We have
   
0
0 0
2
6 11 6 11
6 11
6 11
.
st
st st
t t e dt
te dt e dt
s s


 
 
    
  
  

 
L
2. Use the Laplace transform to solve the following initial value problem.

   '' ' 6 6 11; 0 1; ' 0 4.y y y t y y       

We have
   
2
'' ' 6 6 11
6 11
.
y y y t
s s
    
  
L L
Now we have
       
           
         
      
   
   
2
2
2
2
2
2
'' ' 6 '' ' 6
0 ' 0 0 6
6 1 0 ' 0
6 1 1 4
6 1 4
6 5
6 11
.
y y y y y y
s Y s sy y sY s y Y s
s s Y s s y y
s s Y s s
s s Y s s
s s Y s s
s s
    
       
     
      
     
    
  
L L L L
Thus we have
 
  
  
  
 
2 2
3 2
2
2
2
2
2
2
1 11 6
5
6
5 11 6
.
3 2
2 7 3
3 2
7 3
3
,
3
Y s s
s s s s
s s s
s s s
s s s
s s s
s s
s s
A B C
s s s
 
     
   
  
 
 
  
 
 
  


  

where A, B, and C are constants. Thus we have
   
   
   
2 2
2 2
2
7 3 3 3
3 3
3 3 .
s s As s B s Cs
A s s B s Cs
A C s A B s B
       
    
     
Equating co
esponding coefficients, we obtain the following linear system:
1;
3 7;
3 3.
A C
A B
B
  
  
  
The last equation yields
1.B 

Plugging this value into the second equation, we obtain
3 1 7,A  
whence
3 6A 

and hence
2.A 
Finally, plugging this value of A into the first equation, we obtain
2 1,C   
whence
1.C 

Thus we have
  2
2 1 1
.
3
Y s
s s s
   

Taking the inverse Laplace transform of  Y s , we obtain

  32 .ty t t e   
3. For each of the following boundary value problems, either show the problem has no
solution, find its unique solution, or find all of its infinitely many solutions.
(a)    2 2'' 1; 0 1; .y y x y y
     
First we find the general solution to the above differential equation. We have
,H Py y y 
where Hy is the general solution to the co
esponding homogeneous differential
equation and Py is a particular...
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