- #1

jbord39

- 74

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## Homework Statement

(D^3 - D^2 + D - 1)y = e^x

and y(0) = 0, y'(0) = 1, y"(0) = 0

## Homework Equations

D = d/dx

## The Attempt at a Solution

Factoring the left side of the equation gives :

(D^2 + 1)(D - 1)y = e^x

Which has roots of +/- i and 1.

So y(c) = Ae^x + Bsin(x) + Ccos(x)

The annihilator for e^x is (D-1)

and y(p) = Dxe^x = e^x[Dx]

So y(p)' = D[xe^x + e^x] = e^x[D(x+1)]

y(p)" = D[xe^x + 2e^x] = e^x[D(x+2)]

y(p)"' = D[xe^x + 3e^x] = e^x[D(x+3)]

Plugging into y"' - y" + y' -y = e^x

and dividing by e^x yields:

2D = 1 ; or D = (1/2)

So now y = Ae^x + Bsin(x) + Ccos(x) + (1/2)xe^x

Plugging into initial conditions:

y(0) = A + C = 0 ; OR A = -C

Now y' = Ae^x + Bcos(x) - Csin(x) + (1/2)(xe^x + e^x)

and y'(0) = 1 = A + B + (1/2)(0 + 1)

or A + B = (1/2)

Now y" = Ae^x - Bsin(x) - Ccos(x) + (1/2)(xe^x + 2e^x)

and y"(0) = 0 = A - C + (1/2)(0 + 2)

or A - C + 1 = 0

Solving for A, B, and C yields:

A = (-1/2)

B = 1

C = (1/2)

The solution is therefore:

y = (-1/2)e^x + sin(x) + (1/2)cos(x) + (1/2)xe^x

Does this look correct?

Thanks a bunch