Solving a Second-Order ODE with Non-Real Roots

[Tom1]

Homework Statement

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

Homework Equations

Included in 1

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

 

[Hootenanny]

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.

 

[Tom1]

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

 

[Hootenanny]

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

No problem, what are your two solutions?

 

[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?

 

[Hootenanny]

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?

 

[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)

?

 

[Hootenanny]

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,

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

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?

 

[Tom1]

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

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

 

[Hootenanny]

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,

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

 

[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?

 

[Hootenanny]

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?

 

[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?

 

[Hootenanny]

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

 

[Tom1]

Thank you much for all your help.

 

[Hootenanny]

Thank you much for all your help.

No problem, it was a pleasure