Solving a coupled nonhomegenoues ode

[dawgs1236]

Homework Statement

$\dot{ω_{1}}$ = λ$ω_{2}$ +μ
$\dot{ω_{2}}$ = -λ$ω_{1}$

Homework Equations

λ and μ are real, positive constants
$ω_{1}$(0) ≠ 0
$ω_{2}$(0) ≠ 0

The Attempt at a Solution

I know that the general solution will be in the form
ω1(t) = A sin ωt + B cos ωt + C
ω2(t) = D sin ωt + E cos ωt
but I'm not sure how to solve it

Eqn 2 becomes:
$\ddot{ω_{2}}$ = -λ$\dot{ω_{1}}$
Substituting equation 1:
$\ddot{ω_{2}}$ = -λ$^{2}$$ω_{2}$ +λμ

I'm really just unsure of how to handle the constant when you solve it as a nonhomegenous equation.

 

[Dick]

Homework Statement

$\dot{ω_{1}}$ = λ$ω_{2}$ +μ
$\dot{ω_{2}}$ = -λ$ω_{1}$

Homework Equations

λ and μ are real, positive constants
$ω_{1}$(0) ≠ 0
$ω_{2}$(0) ≠ 0

The Attempt at a Solution

I know that the general solution will be in the form
ω1(t) = A sin ωt + B cos ωt + C
ω2(t) = D sin ωt + E cos ωt
but I'm not sure how to solve it

Take another derivative of the first equation and substitute the second one into it.

 

[dawgs1236]

Yeah I know how to do that part. I'm really just not sure what to do about the μ on the end. When I take the derivative it should disappear but I don't think it should.

 

[dextercioby]

Well, if it's a constant, then it should...

You can of course take the derivative of the second equation and use the first.

 

[dawgs1236]

So does μ affect this problem?

the answers I came up with were

$ω_{1}$(t) = $ω_{2}$(0) sin ωt + $ω_{1}$(0) cos ωt
$ω_{2}$(t) = $ω_{1}$(0) sin ωt + $ω_{2}$(0) cos ωt

but these were wrong

 

[dextercioby]

Of course they're wrong, there's no evidence of mu and lambda.

 

[Dick]

Yeah I know how to do that part. I'm really just not sure what to do about the μ on the end. When I take the derivative it should disappear but I don't think it should.

I think the μ will disappear when you take the derivative. That's ok. Just keep working. It will come back into the solution later. Show your work.

 

[dawgs1236]

Of course they're wrong, there's no evidence of mu and lambda.

Obviously I know they're wrong. Why else would I post this here? I posted it here looking for help with the solution not conformation that I was incorrect.

 

[Dick]

Obviously I know they're wrong. Why else would I post this here? I posted it here looking for help with the solution not conformation that I was incorrect.

Can't say why you got incorrect solutions until you show how you got them.

 

[dawgs1236]

Would it be a good idea to take the second derivative of :
ω2(t) = D sin ωt + E cos ωt
and plug them them into:
ω2¨ = -λ^2ω2 +λμ
and then use the initial conditions?

 

[Dick]

Would it be a good idea to take the second derivative of :
ω2(t) = D sin ωt + E cos ωt
and plug them them into:
ω2¨ = -λ^2ω2 +λμ
and then use the initial conditions?

It would be a good idea to say how you got ω2(t) = D sin ωt + E cos ωt. It looks like you are just plugging in a guess for the answer. And the guess is wrong. Why don't you start with the part we agreed on? Solve (ω1)''=(-λ^2ω1).

 

[dawgs1236]

I did just guess at that.

okay so I set up the characteristic equation and got sqrt(λ) as the answer. would that make it:
ω1(t) = A sin sqrt(λ)t + B cos sqrt(λ)t + C ?

 

[Dick]

I did just guess at that.

okay so I set up the characteristic equation and got sqrt(λ) as the answer. would that make it:
ω1(t) = A sin sqrt(λ)t + B cos sqrt(λ)t + C ?

Good, you're getting started. Try to check that, plug it into (ω1)''=(-λ^2ω1). You should realize the the sqrt(λ) isn't quite right. I get a characteristic equation of x+λ^2=0. And does the C really belong there?

 

[dawgs1236]

Okay so ω=-λ
working that out I came up with:

ω1(t) = w1(0) cos -λt - [w2(0) + μ/λ] sin -λt
ω2(t) = w1(0) sin -λt + [w2(0) + μ/λ] cos (-λt) - μ/λ

 

[Dick]

Okay so ω=-λ
working that out I came up with:

ω1(t) = w1(0) cos -λt - [w2(0) + μ/λ] sin -λt
ω2(t) = w1(0) sin -λt + [w2(0) + μ/λ] cos (-λt) - μ/λ

That works. It would maybe look a little nicer if you would get rid of the signs on λ using trig, but it works. You've certainly gotten a lot better at this since your last post.