Linear differential equation

A, B, a are constants. normally it would be written as

x = Ae^-0.5t + Be^-2t + C

where A, B, C (or C1, C2, C3) are the constants.
you can determine their value if you know initial conditions.
to solve for n unknowns you need n initial conditions so in this case you would need 3.

back to problem:

when solving differential equations, you need to do it in steps.
first you solve HOMOGENEOUS equation (which is what you did).

this is equation in which driving function is zero so in your case homogeneous equation is
2(d²x/dt²) + 5(dx/dt) + 2x =0

it defines natural response of the system (of system whose response is represented by that equation)

then you need to determine particular solutions.
when selecting possible candidates for particular solution, we need to look at function that is on the right side of equal signs as well as existing homogeneous solution (so that THOSE solutions are not repeated).

if forcing function is sin or cos, you have to include BOTH sin and cos in your particular solution.

for example since you have 5sin(t)

we would assume that possible candidate solution is

xp= D*sin(t)+E*cos(t)+F

to test if this would work, we need to plug it into original equation (so we need to find x' and x'' of this 'candidate').

for example
xp= D*sin(t)+E*cos(t)+F
xp'=D*cos(t)-E*sin(t)
xp''=-D*sin(t)-E*cos(t)

and then plug it in and solve:
2(yp'') + 5(yp)+2x=5sin(t)
2(D*cos(t)-E*sin(t)) + 5(-D*sin(t)-E*cos(t))+2(D*sin(t)+E*cos(t…