Alright MrAl, maybe I got this completed. Sorry it took so long, I am certainly no mathematician.
You got me thinking with not sticking to one method...
This particular equation apparently requires Laplacian transforms.
I tried working it using the: dy/dx + P(x) = Q(x) method with an IF, and ended up with: 0.1e^0.3x + C. Which that may not even be the correct answer to "that" method, but either way, I didn't have NEARlY enough constants in the answer so I stopped there.
Ok, here is what I came up with using a Laplace transform:
2sin8t = 10Dx(t) + 3Dx(t), x(0) =7
divide all by 10, to isolate the highest order derivative operator and get: 1/5sin 8t = Dx(t) + 3/10x(t)
*L = Laplace operator
L{1/5 sin8t} = L{d/dt x(t)} + L{3/10 x(t)}
1/5 (8/s² + 8²) = sX(s) - x(0) + 3/10 X(s)
1/5 (8/s² + 8²) = sX(s) - 7 + 3/10 X(s)
1/5 (8/s² + 8²) + 7 = sX(s) + 3/10X(s) = (s+3/10)X(s)
X(s) = 1/5 (8/s² + 8²) + 7 / (s+3/10) = (8/5) / ((s^2 + 8^2)(s+3/10)) + 7/(s+3/10)
Now inverse Laplace:
L^-1 {X(s)} = L^-1{ (8/5) ÷ ((s² + 8²)(s+3/10)) + 7/(s+3/10) }
where: (s² + 8²) = (s+j8)(s-j8)
and the poles are: (s+j8)(s-j8)(s+3/10)
Part. Frac. Decomp.:
A1/(s+j8) + A2/(s-j8) + A3/(s+3/10) + B/(s+3/10) =
(8/5) ÷ (s+j8)(s-j8)(s+3/10)
Do a lot a canceling and setting s = N, where N equals a number to cancel your 2 other A's, you know the drill......
Do Laplace again to get back to the time domain for every A(n) and then finally for your B:
ANSWER ===>
x(t) = (8/5) [ A1 (e^-j8t x e^-3/10t) + A2(e^j8t x e^3/10t) + A3(e^-j8t x e^j8t) + 7e^3/10t)]
x(t) = (8/5) [
A1 (e^-j8t x e^-3/10t) + A2(e^j8t x e^3/10t) +
(A3 + 7e^3/10t)]
Blue = constant sin f(x), and red = transient which eventually goes away.
FINALLY!
Thanks MrAl, and Speyman for the help!