Simple second-order ODE

1. Mar 28, 2008

Tom1

1. The problem statement, all variables and given/known data

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

2. Relevant 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

2. Mar 28, 2008

Hootenanny

Staff Emeritus
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.

3. Mar 28, 2008

Tom1

I have never done one with complex numbers before. Mind walking me through this one?

4. Mar 28, 2008

Hootenanny

Staff Emeritus
No problem, what are your two solutions?

5. Mar 28, 2008

Tom1

Assuming a=1, b=2 c=2 and r1= $$\lambda$$+i$$\mu$$ r2=$$\lambda$$-i$$\mu$$

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

Correct?

6. Mar 28, 2008

Hootenanny

Staff Emeritus
I have different solutions. In your solution, you make the error of assuming that,

$$\sqrt{4} = 2\sqrt{2}$$

Which is not the case. Using the quadratic formula,

$$r = \frac{-2\pm\sqrt{4-4\cdot1\cdot2}}{2} = \frac{-2\pm i\sqrt{4}}{2}$$

$$r = -1\pm i$$

Do you agree?

7. Mar 28, 2008

Tom1

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)

?

8. Mar 28, 2008

Hootenanny

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

$$y(t) = A\exp\left\{t\left(i-1\right)\right\} + B\exp\left\{-t\left(i+1\right)\right\}$$

Which may be re-written,

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

Now, can you write the two complex exponentials in terms of sines and cosine?

Last edited: Mar 28, 2008
9. Mar 28, 2008

Tom1

Ah, I think I see my mistake. If u=1 then:

y=(Ae^-t)cos(t)+(Be^-t)sin(-t)

10. Mar 28, 2008

Hootenanny

Staff Emeritus
Almost there, note that,

$$\sin\left(-\theta\right) = -\sin\theta$$

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,

$$y(t) = e^{-t}C\cos t + e^{-t}D\sin t$$

11. Mar 28, 2008

Tom1

Just figured out how to use the tex feature to make it easier on your eyes. I'm a little confused about the variables:

$$y(t) = Ae^{-t}C\cos t - Be^{-t}D\sin t$$

Is that the solution?

12. Mar 28, 2008

Hootenanny

Staff Emeritus
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,

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

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

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

Collecting the coefficients of sine and cosine,

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

Let us now define two new constants such that,

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

Hence, the general solution,

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

Do you follow?

Last edited: Mar 28, 2008
13. Mar 28, 2008

Tom1

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?

14. Mar 28, 2008

Hootenanny

Staff Emeritus
Correct indeed

15. Mar 28, 2008

Tom1

Thank you much for all your help.

16. Mar 28, 2008

Hootenanny

Staff Emeritus
No problem, it was a pleasure