Initial value nonhomogeneous DE

[jbord39]

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

 

[jbord39]

Anyone know if I'm in the right direction?

 

[jbord39]

Bump.

 

[Dick]

It looks fine. You know you can check your own work by taking the y you get and substituting back into the original ODE? That is satisfies the boundary conditions isn't hard to check either?

 

[jbord39]

It looks fine. You know you can check your own work by taking the y you get and substituting back into the original ODE? That is satisfies the boundary conditions isn't hard to check either?

Thanks, it makes sense but I did not think of that.