Non Homogenous Differential Equation

[shards5]

Homework Statement

y"' - 9y" +18y' = 30e^x
y(0) = 16
y'(0) = 14
y"(0) = 11

Homework Equations

n/a

The Attempt at a Solution

Factor Out
r(r² - 9r +18)
r = 0; r = 6; r =3
General Equation
y(x) = c₀ + c₁e^3x + c₂e^6x
y'(x) = 3c₁e^3x + 6c₂e^6x
y"(x) = 9c₁e^3x + 36c₂e^6x
c₁ = (11 - 36c₂)/9
y'(x) = 14 = 3c₁e^3x + 6c₂e^6x
y'(x) = 14 = ((11 - 36c₂)/9)e^3x + 6c₂e^6x
c₂ = -31/30
c₁ = (11 - 36*(-31/30))/9 = 5.35555556
y(0) = 16 = c₀ + c₁e^3x + c₂e^6x
16 +31/30 - 5.35555556 = c₀ = 11.6777778
Solve For A in Ae^x
y_p = Ae^x
y'_p = Ae^x
y"_p = Ae^x
y"'_p = Ae^x
Inputting into the original equation we get.
Ae^x - 9Ae^x +18Ae^x = 30e^x
Simplifying we get.
10Ae^x = 30e^x
Which gives A = 3 and since 3 is a root of the original equation we add an x to differentiate between the two.
So the final equation SHOULD BE 3xe^x + 11.6777778 + 5.35555556e^3x -31/30e^6x but of course its not.
So my question is, what am I doing wrong?

 

[Mark44]

Most of your work is fine, but I believe you have made an error in your calculations for c_0, c_1, and c_2. I used matrix methods to solve for these constants and got c_2 = 25/18. I'm fairly confident of this value, but didn't check it.

Also, in your last paragraph you say something that isn't true. You got A = 3, which means that your particular solution is y_p = 3e^x. The fact that you got a value of 3 when you solved for A is irrelevant to anything else in this problem. Your general solution will include 3e^x, not 3xe^x.

 

[shards5]

When you say matrix method do you just make a matrix like this?

1 1 1 16
0 3 6 14
0 9 36 11

And then you row reduce?

 

[Mark44]

When you say matrix method do you just make a matrix like this?

1 1 1 16
0 3 6 14
0 9 36 11

And then you row reduce?

Yes, and yes.