Differential Equations Question please help?

Change the equation

y''' - y' = t to y'' - y = t²/2 + C

The characteristic equation is

r²- 1 = 0

r = 1, -1

So the two linearly independent functions to the homogeneous solution is

yp = Av(t)eᵗ + Bw(t)e⁻ᵗ

So our restrictions are

v'(t)eᵗ + w'(t)e⁻ᵗ = 0

v'(t)eᵗ - w'(t)e⁻ᵗ = t²/2 + C

Eliminate by adding. v'(t)eᵗ + w'(t)e⁻ᵗ + v'(t)eᵗ - w'(t)e⁻ᵗ = 2v'(t)eᵗ = t²/2 + C

v'(t) = t²e⁻ᵗ/4 + Ce⁻ᵗ =

v(t) = -¼ e⁻ᵗ[t²+ 2t +2] + Ce⁻ᵗ = (no constant because we only want one particular solution or you could say we took c = 0)

Substitute v'(t) back and w'(t) = Ceᵗ - t²eᵗ/4, so w(t) = Keᵗ -¼eᵗ[t²-2t+2] (same argument for v(t)).

Put it all back together into y = v(t)eᵗ + w(t)e⁻ᵗ

yp = [-¼ e⁻ᵗ[t²+ 2t +2] + Ce⁻ᵗ]eᵗ + [-¼eᵗ[t²-2t+2] - Keᵗ ]e⁻ᵗ

Distribute the e⁻ᵗ and eᵗ to make things simpler. Solving you should get

yp = (C - K) - t²/2 = D - t²/2 (let D = C - K)

Add the homogeneous solution yh = Aeᵗ + Be⁻ᵗ to get general solution

Hence the general solution is

y(t) = Aeᵗ + Be⁻ᵗ + D - t²/2

Yin

I do not know that method, but I am sure it is not the simplest way.

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 = Dt² + Et

yᵖ' = 2Dt + E

yᵖ'' = 2D

yᵖ''' = 0

yᵖ''' - yᵖ' = t

-(2Dt + E) = t

-2Dt - E = t

-2D = 1

2D = -1

D = -½

E = 0

yᵖ = -t² / 2

Find the general solution by combining these two parts:

y = yᶜ + yᵖ

y = A + B℮ᵗ + C / ℮ᵗ - t / 2

y = A - t² / 2 + B℮ᵗ + C / ℮ᵗ

if you thank me, i will give you :you are welcome:

however if you have a question like this please ask from your maths teacher. don't be a foolish to others!!!!!