Non Homogenous Differential Equation

  • Thread starter shards5
  • Start date
  • #1
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Homework Statement


y"' - 9y" +18y' = 30ex
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 + c1e3x + c2e6x
y'(x) = 3c1e3x + 6c2e6x
y"(x) = 9c1e3x + 36c2e6x
c1 = (11 - 36c2)/9
y'(x) = 14 = 3c1e3x + 6c2e6x
y'(x) = 14 = ((11 - 36c2)/9)e3x + 6c2e6x
c2 = -31/30
c1 = (11 - 36*(-31/30))/9 = 5.35555556
y(0) = 16 = c0 + c1e3x + c2e6x
16 +31/30 - 5.35555556 = c0 = 11.6777778
Solve For A in Aex
yp = Aex
y'p = Aex
y"p = Aex
y"'p = Aex
Inputting into the original equation we get.
Aex - 9Aex +18Aex = 30ex
Simplifying we get.
10Aex = 30ex
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 3xex + 11.6777778 + 5.35555556e3x -31/30e6x but of course its not.
So my question is, what am I doing wrong?
 
Last edited:

Answers and Replies

  • #2
35,285
7,131
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.
 
  • #3
38
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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?
 
  • #4
35,285
7,131
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.
 

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