Verify unsolvable ODE on Midterm

[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

 

[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.

 

[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?

 

[gabbagabbahey]

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

 

[gr3g1]

Hmm...

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

 

[gabbagabbahey]

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

 

[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?

 

[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?

 

[gr3g1]

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

 

[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?

 

[gr3g1]

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

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

 

[gabbagabbahey]

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

 

[gr3g1]

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

 

[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

 

[HallsofIvy]

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!