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Simple second-order ODE

  • Thread starter Tom1
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1. Homework Statement

Solve the initial value problem:
y''+2y'+2y=0
y([tex]\pi[/tex]/4)=2
y'([tex]\pi[/tex]/4)=-2

2. Homework Equations

Included in 1

3. The Attempt at a Solution

I assumed that y=e^rt and came up with the characteristic polynomial r^2+2r+2=0 but when solving it for the two roots, they come back as non-real. Where have I gone wrong? thanks
 

Answers and Replies

Hootenanny
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You've done nothing wrong, the solutions to ODE's may be complex. However, one may transform them into real solutions using Euler's identity.
 
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I have never done one with complex numbers before. Mind walking me through this one?
 
Hootenanny
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I have never done one with complex numbers before. Mind walking me through this one?
No problem, what are your two solutions?
 
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Assuming a=1, b=2 c=2 and r1= [tex]\lambda[/tex]+i[tex]\mu[/tex] r2=[tex]\lambda[/tex]-i[tex]\mu[/tex]

r1=(-1+(-4)^1/2)/2
= -(1/2)+i(2)^(1/2)
r2 = -(1/2)-i(2)^(1/2)

Correct?
 
Hootenanny
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I have different solutions. In your solution, you make the error of assuming that,

[tex]\sqrt{4} = 2\sqrt{2}[/tex]

Which is not the case. Using the quadratic formula,

[tex]r = \frac{-2\pm\sqrt{4-4\cdot1\cdot2}}{2} = \frac{-2\pm i\sqrt{4}}{2}[/tex]

[tex]r = -1\pm i[/tex]

Do you agree?
 
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Indeed, that's a mistake I commonly make.

So with those solutions, the general solution of the IVP would be:

y=c1(e^-t)cos(1)+c2(e^-t)sin(-1)

?
 
Hootenanny
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Indeed, that's a mistake I commonly make.

So with those solutions, the general solution of the IVP would be:

y=c1(e^-t)cos(1)+c2(e^-t)sin(-1)

?
I think that you've made a mistake somewhere (perhaps missing a t here and there...?). Let's start from the top,

[tex]y(t) = A\exp\left\{t\left(i-1\right)\right\} + B\exp\left\{-t\left(i+1\right)\right\}[/tex]

Which may be re-written,

[tex]y(t) = e^{-t}\left[Ae^{it}+Be^{-it}\right][/tex]

Now, can you write the two complex exponentials in terms of sines and cosine?
 
Last edited:
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Ah, I think I see my mistake. If u=1 then:

y=(Ae^-t)cos(t)+(Be^-t)sin(-t)
 
Hootenanny
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Almost there, note that,

[tex]\sin\left(-\theta\right) = -\sin\theta[/tex]

Furthermore, one must not use the same constants (if you work through the full solution you will see why). Hence, the general solution will be of the form,

[tex]y(t) = e^{-t}C\cos t + e^{-t}D\sin t[/tex]
 
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Just figured out how to use the tex feature to make it easier on your eyes. I'm a little confused about the variables:

[tex]y(t) = Ae^{-t}C\cos t - Be^{-t}D\sin t[/tex]

Is that the solution?
 
Hootenanny
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I tell you what, since it is clear that you've put the effort in, I'll provide a complete solution in this case. Starting from the second equation in post #8 one may write,

[tex]y(t) = e^{-t}\left[A\left\{\cos t + i\sin t\right\}+B\left\{\cos\left(-t\right) + i\sin\left(-t\right)\right\}\right][/tex]

Noting that cosine and sine are even and odd function respectively, one may write,

[tex]y(t) = e^{-t}\left[A\left\{\cos t + i\sin t\right\}+B\left\{\cos t - i\sin t\right\}\right][/tex]

Collecting the coefficients of sine and cosine,

[tex]y(t) = e^{-t}\left[\left(A+B\right)\cos t + \left(A-B\right)i\sin t\right][/tex]

Let us now define two new constants such that,

[tex] C:= A+B \hspace{2cm} D:=\left(A-B\right)i \hspace{2cm}C,D\in\mathbb{R}[/tex]

Hence, the general solution,

[tex]y(t) = e^{-t}\left[C\cos t + D\sin t\right][/tex]

Do you follow?
 
Last edited:
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Alright, I see now...but it causes another question to arise: When I go back to find the constants, I apply the initial conditions to solve for C&D, correct?
 
Hootenanny
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Alright, I see now...but it causes another question to arise: When I go back to find the constants, I apply the initial conditions to solve for C&D, correct?
Correct indeed :approve:
 
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Thank you much for all your help.
 
Hootenanny
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Thank you much for all your help.
No problem, it was a pleasure :smile:
 

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