Question

Find the general solution of the given differential equation.
Y^(4prime)-2y (double prime)-8y=e^(-t)+3.
I got r=2,-2, I (-2)^(1/2), I (2)^(1/2). I need help finding Y-p

Answer 1

I think that is suppose to be
\[r=2,-2,- \sqrt{2}i , \sqrt{2}i\]
ok anyways i guess you now know the homogeneous solution
to find the particular you need
y=Ae^(-t)+B
read these notes http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
anyways we could also try the particular solution
y=Ae^(-t)+B+Ct
we should get C is O but you can see what happens if you want

Answer 2

I got y=Ate^(-t)+Bt+C. Is this correct?
And the general solution for the homogeneous part is y=c1e^(2t)+c2^(-2t). I don't know how to include r3 and r4 in the solution.

Answer 3

\[\text{ if } r=a+bi \text{ then we have } y=c_1e^{(a+bi)t}=c_1 e^{at}e^{bt i}=c_1 e^{at}(\cos(bt)+i \sin(bt)) \\ \text{ though this is mostly written as } y=c_1 e^{at}\cos(bt)+c_2 e^{at}\sin(bt)\]

Answer 4

so your homogeneous solution should look like
\[y_h=c_1e^{2t}+c_2e^{-2t}+c_3 \sin( \sqrt{2} t)+c_4 \cos(\sqrt{2} t)\]

Answer 5

\[\sin(-x)=-\sin(x) \\ \text{ and } \cos(-x)=\cos(x)\]
so we don't need to put +\[c_5 \sin(-\sqrt{2}t)+c_6\cos(-\sqrt{2}t)\]
since that is equal to \[-c_5 \sin( \sqrt{2} t)+c_6 \cos(\sqrt{2} t)\]

Answer 6

your particular solution will work
but plug into find A, B , and C

Answer 7

You are a genius

Answer 8

so if you have
\[r=a \pm bi \text{ then } y=c_1 e^{at}\cos(bt)+c_2e^{at}\sin(bt)\]

Answer 9

but anyways your welcome

Answer 10

I posted another question earlier but had to close it in order to post this one. I don't know if you still can see it. I didn't get an answer for it