# This is review for me, but something is going awry

1. Apr 22, 2010

### Jamin2112

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

Find the solution of the following initial value problems:

(a) y'' - 2y' + 5y = 0, y(π/2)=0, y'(π/2)=0

2. Relevant equations

Just know to "guess" y=Cert, how to bring in Euler's formula, and that sin(x) is odd while cos(x) is even

3. The attempt at a solution

Guess: y= ert

-----> r2ert - 2rert + 5ert = 0
-----> r2 - 2r + 5 = 0
-----> (r-1)2 = -5 + 1
-----> r - 1 = +/- √-4
-----> r = 1 +/- 2i
-----> y = et( C1(cos2t+isin2t) + C2(cos(-2t)+isin(-2t)) )
-----> y = et( C1cos2t + C1sin2t - C2cos2t + C2isin2t)
-----> y = et(K1cos2t + K2sin2t), where K1, K2 are new constants to represent (C1 - C2), (iC1 + iC2)

But then come the initial conditions!

0 = eπ/2K1(-1)
2 = K1eπ/2(-1)+ 2K2eπ/2(-1)

It just seems like a ghey problem if K1 disappears

Last edited: Apr 22, 2010
2. Apr 22, 2010

### Dick

Looks good so far. Did you mean y'(pi/2)=2? What's so hard about the initial conditions? What are sin(pi) and cos(pi)?

3. Apr 22, 2010

### Jamin2112

Oh, and another one.

y'' + 2y' + y = 3e-t

I'll find the homogenous solution first. The characteristic polynomial is r2 + 2r + 1 = 0 -----> r = -1, a repeated root ----> The homogenous solution is (C1 + C2t)e-t.

Now I'll guess a second general solution, C3e-t.

------> C3et - 2C3e-t + C3e-t = 3e-t
------> 0 = 0.

Fork! So is the homogeneous solution all? Just for the heck of it, I tried y(t)=(At2 + Bt + C)e-t, but to no avail; there doesn't seem to be a polynomial that can be multiplied by e-t without my getting 0=0.

Where should I go from here?

Last edited: Apr 22, 2010
4. Apr 22, 2010

### Dick

There is an inhomogeneous solution of the form (At^2+Bt+C)*exp(-t). You just missed it.

5. Apr 22, 2010

### Jamin2112

Well, yeah.

y(t) = C1etcos2t + C2etsin2t -----> y'(t) = et(-C12sin2t + C1cos2t + 2C2cost + C2sin2t).

So,
0 = C1eπ/2cos(π) + C2eπ/2sin(π)
2 = -2C1eπ/2sin(π) + C1eπ/2cos(π) + 2C2eπ/2cos(π) + C2eπ/2sin(π)

----->

2 = -C2eπ/2

----> y(t) = -e-π/2etcos2t

Last edited: Apr 22, 2010
6. Apr 22, 2010

### Jamin2112

Allow me to demonstrate.

y(t) = At2e-t + Bte-t + Ce-t
----> y'(t)= -At2e-t + 2Ate-t -Bte-t + Be-t - Ce-t

----> y''(t)= At2e-t - 2Ate-t - 2Ate-t + 2Ae-t + Bte-t -Be-t - Be-t + Ce-t

-----> At2e-t - 2Ate-t - 2Ate-t + 2Ae-t + Bte-t -Be-t - Be-t + Ce-t + 2*[ -At2e-t + 2Ate-t -Bte-t + Be-t - Ce-t ] + At2e-t + Bte-t + Ce-t = 3e-t

Everything on the left side cancels out!

7. Apr 22, 2010

### Dick

Yes, what's wrong or 'ghey' about that?

8. Apr 22, 2010

### Jamin2112

I'm used to easy system of equations in these problems.

Something like

0 = C1 + C2
2 = 3C1 - C2,

not something with eπ/2

9. Apr 22, 2010

### Dick

Look, you can forget about the B and C parts. You should KNOW they are going to cancel. They are part of the homogenous solution. Are you really going to make me check that? I will ask you, what cancelled the 2*A*exp(-t)?

10. Apr 22, 2010

### Dick

exp(pi/2) is just another constant. Doesn't make it any harder than a constant like '3'.

11. Apr 22, 2010

### Jamin2112

Let us see.

y(t) = At2e-t,
y'(t) = -At2e-t + 2Ate-t,
y''(t)=At2e-t - 2Ate-t - 2Ate-t + 2Ae-t

------------>

At2e-t - 2Ate-t - 2Ate-t + 2Ae-t -2At2e-t + 4Ate-t + At2e-t = 3e-t

----------> A = 3/2

I think I finally got it. What a ghey problem.

WAIT! Now I have the homogeneous solution, in which the coefficients are undetermined, PLUS one another e-t with a determined coefficient. How does that work? What was the point if (3/2 + C1) could just be merged into a new constant?

Last edited: Apr 22, 2010
12. Apr 22, 2010

### Dick

I would stop saying 'ghey'. I don't think it's universally appreciated. You can't absorb the inhomogeneous solution into the homogeneous solution by adjusting a constant. You are just being sloppy again.