DE constant coefficients and boundary conditions

  • Thread starter AkilMAI
  • Start date
  • #1
62
0

Homework Statement


Find the solution of the equation
v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159


Homework Equations


........



The Attempt at a Solution


I treat it as a polynomial=>r^2+4r+5=0
=>delta=-4=>r1=2+2i and r2=2-2i
v=e^[x+2](A*cos[2]+B*i*sin[2])
v=-1=e^[pi+2](A*cos[2]+B*i*sin[2])
v'=0....so I"m stuck and I need some help
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
963

Homework Statement


Find the solution of the equation
v''- 4v'+5v=0,such that v=-1 and v'=1 when x=pi=3.14159


Homework Equations


........



The Attempt at a Solution


I treat it as a polynomial=>r^2+4r+5=0
This is NOT the characteristic equation of the d.e. you posted. Look at your original problem again.

=>delta=-4=>r1=2+2i and r2=2-2i
v=e^[x+2](A*cos[2]+B*i*sin[2])
Not "cos(2)", cos(2x)!

v=-1=e^[pi+2](A*cos[2]+B*i*sin[2])
v'=0....so I"m stuck and I need some help
 
  • #3
62
0
yes you are right,now I get this system
e^[pi+2](A*cos[2x]+B*i*sin[2x])=-1
2e^[pi+2](B*i*cos[2x]-A*sin[2x])=1...how can I solve it?
 
  • #4
dextercioby
Science Advisor
Homework Helper
Insights Author
13,058
606
Doesn't look ok. You shouldn't have any x left in the 2 equations. You need to plug x=pi everywhere. That will simplify your system.
 
  • #5
62
0
ok...
e^[pi+2](A*cos[2pi]+B*i*sin[2x])=-1=>e^[pi+2]*B*i=-1
2e^[pi+2](B*i*cos[2pi]-A*sin[2x])=1.=>2e^[pi+2]*A=1
how can I find A and B?
 
  • #6
dextercioby
Science Advisor
Homework Helper
Insights Author
13,058
606
Well, the ODE is [itex] v''-4v'+5v =0 [/itex]. The general solution I found is

[tex] v(x)= Ae^{(2+i)x} + Be^{(2-i)x} [/tex]

Now you'll get 2 equations from

[tex] v(\pi) = -1 [/tex]
[tex] v'(\pi) = 1 [/tex]

Can you spell them out again ?
 
  • #7
62
0
A*e^((i+2)pi)+B*e^((i+2)pi)=-1
A*(i+2)*e^((i+2)pi)+B*(i-2)*e^(-(i-2)pi)=1........???
 
  • #8
Dick
Science Advisor
Homework Helper
26,260
619
A*e^((i+2)pi)+B*e^((i+2)pi)=-1
A*(i+2)*e^((i+2)pi)+B*(i-2)*e^(-(i-2)pi)=1........???

This is getting to be a mess. Let's start over again. The equation is v''- 4v'+5v=0, right? What are the roots of the associated quadratic?
 
  • #9
62
0
2+2i and 2-2i
 
  • #10
62
0
sorry 2+i and 2-i
 
  • #11
Dick
Science Advisor
Homework Helper
26,260
619
2+2i and 2-2i

That's your first problem. The imaginary part of those is wrong. How did you get it?
 
  • #12
62
0
no,It is 2-i and 2+i.....I calculated delta=-4=>r1=2-i and r2=2+i
 
  • #13
Dick
Science Advisor
Homework Helper
26,260
619
Ok, and can you write, for example, A*exp((2-i)*x) in terms of the trig functions?
 
Last edited:
  • #14
62
0
as in, e ^(2*x)*(P*cos( x) + Q*sin( x)),where P = A+B and Q = i(A−B)?
 
  • #15
Dick
Science Advisor
Homework Helper
26,260
619
as in, e ^(2*x)*(P*cos( x) + Q*sin( x)),where P = A+B and Q = i(A−B)?

Yes, that's what I meant. So v=e ^(2*x)*(P*cos( x) + Q*sin( x)). Now you just have to put in your boundary conditions and find P and Q.
 
Last edited:
  • #16
62
0
ok,I'll reply with a result in 5 min
 
  • #17
62
0
e ^(2*pi)*(P*cos( pi) + Q*sin( pi))=-1=>e ^(2*pi)*(-P)=-1
e ^(2*pi)*((2P+Q)*cos( pi) -(P- 2Q)*sin( pi))=1 =>e ^(2*pi)*(-(2P+Q))=1....multiplying ome of the wtih -1 is not going to help me solve the system
 
  • #18
Dick
Science Advisor
Homework Helper
26,260
619
e ^(2*pi)*(P*cos( pi) + Q*sin( pi))=-1=>e ^(2*pi)*(-P)=-1
e ^(2*pi)*((2P+Q)*cos( pi) -(P- 2Q)*sin( pi))=1 =>e ^(2*pi)*(-(2P+Q))=1....multiplying ome of the wtih -1 is not going to help me solve the system

Your first equation tells you what P is. Put that into the second equation.
 
  • #19
62
0
if P=-1/e ^(2*pi)=>Q=1
 
  • #20
Dick
Science Advisor
Homework Helper
26,260
619
if P=-1/e ^(2*pi)=>Q=1

You are being sloppy. What's P? You've got a sign wrong. And Q is way wrong. How did you do that?
 
  • #21
62
0
I know I'm sloppy,is because I'm tired.....P=1/e ^(2*pi)=>-e ^(2*pi)*((2/e ^(2*pi))+Q)=1
 
  • #22
62
0
=>Q=1/e ^(2*pi)-2/e ^(2*pi)
 
  • #23
Dick
Science Advisor
Homework Helper
26,260
619
=>Q=1/e ^(2*pi)-2/e ^(2*pi)

Take a rest. You are missing a sign again. Q=-1/e ^(2*pi)-2/e ^(2*pi). Note the first minus.
 
  • #24
62
0
I will,but first this...so these are P and Q...I plug them into e ^(2*x)*(P*cos( x) + Q*sin( x))...and the result is the solution of the equation.
 
  • #25
62
0
I hope I'm not missing something
 

Related Threads on DE constant coefficients and boundary conditions

  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
896
J
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
979
Replies
1
Views
1K
Replies
1
Views
911
  • Last Post
Replies
1
Views
871
Top