Property of exponentials and sine functions?

[Unsilenced]

Homework Statement

x''(t)+r*x't+kx=0

Suppose that for some initial conditions the solution is given by

x=e^(-2t)*(3cos(t)+4sin(t))

What are are and k?

Homework Equations

See above

The Attempt at a Solution

I've tried to "brute force" the solution simply by sticking the expression for x into the ODE, but that quickly becomes very complicated, and is easy to mess up. I can check my answers with wolfram alpha or some such, but it still will be difficult to pinpoint errors.

Instead, I've heard that the values 3 and 4 don't matter (presumably they are determined by initial conditions, not r and k), and that there is a (relatively) simple expression that will make it easy to get the derivative and double derivative of this expression. I assume Euler's formula is involved somehow, but don't know exactly how to work it so that I can turn the entire function into an exponential.

 

[lmedin02]

3 and 4 do not matter. Work backwards. If the solution in this case involves sines and cosines then we can conclude that the roots to the auxiliary equation are complex. Find the roots and then the polynomial. You will then have your differential equation and hence the values of r and k.

 

[Unsilenced]

3 and 4 do not matter. Work backwards. If the solution in this case involves sines and cosines then we can conclude that the roots to the auxiliary equation are complex. Find the roots and then the polynomial. You will then have your differential equation and hence the values of r and k.

So, I dropped the 3 and the 4 and used Euler's formula to turn them into exponentials.

I ended up with

x=ie^((-2-i)t)-ie^((-2+i)t)

Does that mean my roots are -2 +/- i? Or do I have to go further?Edit: Working through, I got
k=5
r=-4

Seems like a legit answer, but was dropping the i values in front of the exponents a valid move, or should I have gotten something else for my roots?

 

[lmedin02]

No, that is a bit complicated. If you have the differential equation x''-4x'+x=0, how would you find the general solution? Just go through a few steps, then should have the solution to your problem as well. If not, I will give you another hint.

 

[vela]

So, I dropped the 3 and the 4 and used Euler's formula to turn them into exponentials.

I ended up with

x=ie^((-2-i)t)-ie^((-2+i)t)

Does that mean my roots are -2 +/- i? Or do I have to go further?

In your OP, the exponential part was $e^{+2t}$ while your work here implies it's $e^{-2t}$. Which one is correct?

Dropping the 3 and 4 the way you did turns out to work okay to get the answer, but it would be better if you followed Imedin02's suggestion to see why it works out that way.

 

[Unsilenced]

No, that is a bit complicated. If you have the differential equation x''-4x'+x=0, how would you find the general solution? Just go through a few steps, then should have the solution to your problem as well. If not, I will give you another hint.

Characteristic equation: a^2-4a+1=0

Using the quadratic formula gets me a=2+/-sqrt(3)

x=C1e^(2+sqrt3)+C2e^(2-sqrt3)

...

Did I do something wrong?

In your OP, the exponential part was $e^{+2t}$ while your work here implies it's $e^{-2t}$. Which one is correct?

Dropping the 3 and 4 the way you did turns out to work okay to get the answer, but it would be better if you followed Imedin02's suggestion to see why it works out that way.

Um, negative would appear to be correct. Sorry.

 

[lmedin02]

That is correct, however, I made a typo in my DE. It should be x''-4x'+5x=0. The idea is that you can find the characteristic equation from the DE, hence the roots, and then the general solution. So you may work backwards as well. Try it again.

 

[Unsilenced]

That is correct, however, I made a typo in my DE. It should be x''-4x'+5x=0. The idea is that you can find the characteristic equation from the DE, hence the roots, and then the general solution. So you may work backwards as well. Try it again.

Ah, I found where the sign flip came in. Values are now 4 and 5, and the roots what they should be, assuming I had the correct roots to begin with.