# DE constant coefficients and boundary conditions

## Homework Statement

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

........

## 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

HallsofIvy
Homework Helper

## Homework Statement

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

........

## 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

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?

dextercioby
Homework Helper
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.

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?

dextercioby
Homework Helper
Well, the ODE is $v''-4v'+5v =0$. The general solution I found is

$$v(x)= Ae^{(2+i)x} + Be^{(2-i)x}$$

Now you'll get 2 equations from

$$v(\pi) = -1$$
$$v'(\pi) = 1$$

Can you spell them out again ?

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........???

Dick
Homework Helper
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?

2+2i and 2-2i

sorry 2+i and 2-i

Dick
Homework Helper
2+2i and 2-2i

That's your first problem. The imaginary part of those is wrong. How did you get it?

no,It is 2-i and 2+i.....I calculated delta=-4=>r1=2-i and r2=2+i

Dick
Homework Helper
Ok, and can you write, for example, A*exp((2-i)*x) in terms of the trig functions?

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as in, e ^(2*x)*(P*cos( x) + Q*sin( x)),where P = A+B and Q = i(A−B)?

Dick
Homework Helper
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.

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ok,I'll reply with a result in 5 min

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

Dick
Homework Helper
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.

if P=-1/e ^(2*pi)=>Q=1

Dick
Homework Helper
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?

I know I'm sloppy,is because I'm tired.....P=1/e ^(2*pi)=>-e ^(2*pi)*((2/e ^(2*pi))+Q)=1

=>Q=1/e ^(2*pi)-2/e ^(2*pi)

Dick