1. ## undetermined coefficients help

Can someone tell me if this is right? if it isnt, tell me what is wrong. Im lost on creating the correct complimentary and particular solutions..

1. $y'' - 2y' +5y = e^x\cos 2x$

2. $m^2 - 2m + 5 = 0$

3. $m = 1 +2i$

4. $m = 1 -2i$

5. $Yc = e^x(A\cos 2x + B\sin 2x)$

6. $Yp = xe^x(A\cos 2x + B\sin 2x)$

I understand when you have the same answers in the particular solution you have to add an 'x' to the particular solution. What im lost on is getting the correct particular solution.

thanks!

2. Why again do you include the factor x in front of the last solution? You have no multiple roots. Your solution in 5 is the general real solution.

Also, to get a particular solution, you need initial conditions, which you don't seem to have.

3. Well if $Yc = e^x(A\cos2x +B\sin2x)$ Why is it not repeated? When you solve the left side of the equation, you get $Yc = e^x(A\cos2x +B\sin2x)$... but then to solve for the particular solution you would normal use undertermined coefficients. The particular solution would have the form $e^x(A\cos2x + B\sin2x)$ but it is repeated in the complimentary solution isnt it? Thats where im lost.....

4. Hi,
First off, I think we should ignore your term "repeated" for now and also,
to get a particular solution, you need initial conditions, which you don't seem to have
we can get the particular solution without initial conditions.

I think you have 1 thru 5 right.
The way I learned to find the particular solution is to "guess" what a solution might look like, then go thru a bunch of algebra to come out with an answer (I think "guess" is just a laymans term for the method of undetermined coefficienst)

Now for your right hand side you have:
{e}^{x}cos(2x)
Notice that this looks like {e}^{x} times the real part of a complex number, i.e. cos(2x)
It would be very useful to recall Euler's identity here and rewrite the right hand side.

5. what would the right hand side look like then?

6. Euler's identity:
${e}^{i\theta} = cos(\theta) + isin(\theta)$
So rewriting the right hand side we have
${e}^{x}{e}^{i2x} \Rightarrow {e}^{x(1 + 2i)}$
This will be easier to work with rather than using sines and cosines.
So now we have
$y'' - 2y' + 5y = {e}^{x(1 + 2i)}$

Do you know what we do next?

This turned out to be a very tedious exercise, but I hope you will understand how to do it once we are through.