# Verify unsolvable ODE on Midterm

1. Nov 7, 2008

### gr3g1

Long story short,

Professor was replaced with a replacement professor on the day of the midterm.
The replacement prof. announces, after 3/4 of the time allowed has passed that the first question might contain a typo. He doesn't suggest how to fix the problem or anything. He just claims there might be a problem with the first question.

Can someone take a look at it?
This is for Ordinary Differential Equations I

Here is the first question:

Solve the initial value problem:
y' ' ' + y' ' + 4y' + 4 = 0 y (0) = 5; y' (0)=7; y' ' (0)= -5;

Thank you very much

2. Nov 7, 2008

### gabbagabbahey

I don't think there is any typo, it seems easily solvable to me; just use the method of undetermined coefficients, you have a fairly simple 3rd degree inhomogeneous ODE.

3. Nov 7, 2008

### gr3g1

y' ' ' + y' ' + 4y' + 4 = 0
would be come
y' ' ' + y' ' + 4y' = -4

m^3 + m^2 + 4m = 0
m(m^2 + m + 4) = 0

m1 = 0 and m2 = a complex root

is that right?

4. Nov 7, 2008

### gabbagabbahey

That looks like a good start for your complimentary solution.What form does that portion of your solution take?

5. Nov 7, 2008

### gr3g1

Hmm...

e^a(c1cos(bx)+c2sin(bx))

6. Nov 7, 2008

### gabbagabbahey

That's part of it, what are the values of 'a' and 'b', and what happened to your m1=0 root?

7. Nov 7, 2008

### gr3g1

a would be -1/2 and b sqrt(15)/2

We have never seen the case were m1 and m2 are real and complex.. Where would I go from here?

8. Nov 7, 2008

### gabbagabbahey

a real root m1 just adds an c3*e^(m1 x) term, in this case m1=0 and e^0=1, so it adds a constant term and your complimentary solution is:

$$y_c(x)=e^{\frac{-x}{2}} \left( c_1 cos \left( \frac{\sqrt{15}}{2} x \right)+c_2 sin \left(\frac{\sqrt{15}}{2} x \right) \right) +c_3$$

Do you follow?

9. Nov 7, 2008

### gr3g1

Yes, I follow... Where would I go from here?

10. Nov 7, 2008

### gabbagabbahey

Now you need to find a particular solution....Your inhomogeneous term is just '-4'. Suppose you had only a second order ODE, what would you guess as a particular solution there? Can you guess the form of a particular solution for the 3rd order ODE?

11. Nov 7, 2008

### gr3g1

Ok I get it from this point on, however, weve never really had any practice with m1 = 0 and m2 = complex

I wonder why the prof said it was a typo now.. lol..

12. Nov 7, 2008

### gabbagabbahey

maybe because he thought that the sqrt(15)/2 was a little too ugly for an exam question, but who knows :shrug:

13. Nov 7, 2008

### gr3g1

We weren't allowed calculators.. I don't think that changes much though

14. Nov 7, 2008

### gabbagabbahey

My guess is your replacement prof didn't have his morning coffee, and made an error in his attempt to solve the problem, and couldn't find it so he concluded something might b wrong with the question.

But that's just speculation on my part

15. Nov 7, 2008

### HallsofIvy

Staff Emeritus
He probably looked at y' ' ' + y' ' + 4y' + 4 = 0 and thought perhaps it should be
y' ' ' + y' ' + 4y' + 4y = 0 which would be more "standard form" but distinctly harder than the original problem!