Differential Equations Question please help?

The characteristic equation is

r³ - r = 0

r(r²- 1) = 0

r = 0, 1, -1

So the homogeneous solution is

yh = Aeᵗ + Be⁻ᵗ + C

A form to the 2sin(t) would be

yp = Dsin(t) + Ecos(t)

yp + yh = y(t) = Aeᵗ + Be⁻ᵗ + C + Dsin(t) + Ecos(t)

y'(t) = Aeᵗ - Be⁻ᵗ + Dcos(t) - Esin(t)

y''(t) = Aeᵗ + Be⁻ᵗ - Dsin(t) - Ecos(t)

y'''(t) = Aeᵗ - Be⁻ᵗ - Dcos(t) + Esin(t)

We only want y'''(t) and y'(t)

So y'''(t) - y'(t) = Aeᵗ - Be⁻ᵗ - Dcos(t) + Esin(t) - [Aeᵗ - Be⁻ᵗ + Dcos(t) - Esin(t)] = 2Esin(t) - 2Dcos(t)

Matching the coefficients, we get E = 1, D = 0,

Hence the general solution is y(t) = Aeᵗ - Be⁻ᵗ + C + cos(t)

Yin

Solving

http://img18.imageshack.us/img18/2815/473e.gif

Find the complementary function by solving the auxiliary equation:

y''' - y' = 0

m³ - m = 0

m(m² - 1) = 0

m(m + 1)(m - 1) = 0

m = 0 OR m + 1 = 0 OR m - 1 = 0

m = 0 OR m = -1 OR m = 1

yᶜ = A + B℮ᵗ + C℮⁻ᵗ

yᶜ = A + B℮ᵗ + C / ℮ᵗ

Find the particular integral by comparing coefficients:

yᵖ = Dsint + Ecost

yᵖ' = Dcost - Esint

yᵖ'' = -Dsint - Ecost

yᵖ''' = -Dcost + Esint

yᵖ''' - yᵖ' = 2sint

-Dcost + Esint - (Dcost - Esint) = 2sint

-Dcost + Esint - Dcost + Esint = 2sint

2Esint - 2Dcost = 2sint

2E = 2

E = 1

-2D = 0

D = 0

yᵖ = cost

Find the general solution by combining these two parts:

y = yᶜ + yᵖ

y = A + B℮ᵗ + C / ℮ᵗ + cost

Integrate through by t

y'' - y = C - 2cos(t)

Now you have a second order linear homogeneous DE.