- #1
Lyuokdea
- 154
- 0
Ok, this one is really sticking me up:
[tex]y'' - 2y' + 2y = e^{t}cos(t)[/tex]
I solved the homogenous version and got roots 1 +/- i and put these into get the equation
[tex]y_h = c_1e^tcos(t) + c_1e^tsin(t)[/tex]
And I found that the root for e^tcos(t) should be (D- (1 +/- i)
But I'm completely stuck on what to do after this, the part that really confuses me is that in the answer, they only have a e^t sin(t) factor in the answer in addition to the normal homogenous answer, why does sin(t) appear without cos(t) ?
Thanks for your help,
~Lyuokdea
[tex]y'' - 2y' + 2y = e^{t}cos(t)[/tex]
I solved the homogenous version and got roots 1 +/- i and put these into get the equation
[tex]y_h = c_1e^tcos(t) + c_1e^tsin(t)[/tex]
And I found that the root for e^tcos(t) should be (D- (1 +/- i)
But I'm completely stuck on what to do after this, the part that really confuses me is that in the answer, they only have a e^t sin(t) factor in the answer in addition to the normal homogenous answer, why does sin(t) appear without cos(t) ?
Thanks for your help,
~Lyuokdea