Solving a Second-Order ODE with Non-Real Roots

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In summary, the student attempted to solve an IVP but made a mistake. They provided two solutions for the general solution.
  • #1
Tom1
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Homework Statement



Solve the initial value problem:
y''+2y'+2y=0
y([tex]\pi[/tex]/4)=2
y'([tex]\pi[/tex]/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
 
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  • #2
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
I have never done one with complex numbers before. Mind walking me through this one?
 
  • #4
Tom1 said:
I have never done one with complex numbers before. Mind walking me through this one?
No problem, what are your two solutions?
 
  • #5
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?
 
  • #6
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?
 
  • #7
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
Tom1 said:
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?
 
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  • #9
Ah, I think I see my mistake. If u=1 then:

y=(Ae^-t)cos(t)+(Be^-t)sin(-t)
 
  • #10
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]
 
  • #11
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?
 
  • #12
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?
 
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  • #13
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
Tom1 said:
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:
 
  • #15
Thank you much for all your help.
 
  • #16
Tom1 said:
Thank you much for all your help.
No problem, it was a pleasure :smile:
 

1. What is a second-order ordinary differential equation (ODE)?

A second-order ordinary differential equation is a mathematical equation that involves a function and its derivatives up to the second order. It is used to model systems that involve change over time, such as motion, heat transfer, and population growth.

2. How do you solve a simple second-order ODE?

To solve a simple second-order ODE, you can use several methods such as separation of variables, substitution, or the method of undetermined coefficients. The choice of method depends on the specific equation and its initial conditions.

3. What is the difference between a first-order and a second-order ODE?

The main difference between a first-order and a second-order ODE is the number of derivatives involved. A first-order ODE only involves the first derivative of the function, while a second-order ODE involves the first and second derivatives.

4. Can second-order ODEs be used to model real-world phenomena?

Yes, second-order ODEs are widely used in many fields of science and engineering to model real-world phenomena. They can accurately describe the behavior of physical systems and provide insight into their dynamics and behavior.

5. What are some common applications of simple second-order ODEs?

Some common applications of simple second-order ODEs include modeling the motion of objects under the influence of forces, analyzing the behavior of electrical circuits, and predicting the growth or decay of populations.

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