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Tough Diff EQ problem

  • Thread starter hils0005
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[SOLVED] Tough Diff EQ problem

1. Homework Statement
y^(4)-8y^(2)-9y=7e^(-3x)

y^(4)=fourth derivative, y^(2)=second deriv.
2. Homework Equations
Need to find complimentary solution and particular solution using variation of parameters



3. The Attempt at a Solution
r^4-8r^2-9=0
(r^2-9)(r^2+1)=0, r=9,-1

Y(c)=Ce^(9x)+Ce^(-x)

I think the comp soln is right, but I am unsure on how to start on Y(p)
 

Answers and Replies

Hootenanny
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Are you sure about your homogenous solution? I suggest that you re-check your solutions of the indicial equation.
 
Last edited:
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e^(-3x) right?

I tried taking the 4th deriv and 2nd deriv and plugging in which would equal

81e^-3x - 8(9e^-3x) - 9e^(-3x)=7e^(-3x)
but the left side = 0??
 
Hootenanny
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Hootenanny
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I'm quoting from the PM you sent me. As I said via PM, please note that in order to keep our assistance as transparent as possible, it is the policy of Homework Helpers not to provide tuition via PM.
hils0005 said:
sorry hoot,
its been along time since I have had to factor higher polynomials, I think this is correct:
Y(c)=(r-3)(r^3+3r^2+r+3)
(r-3)(r+3)(r^2+1), r=3, -3, 1+i
y(c)=Ce^(3x)+Ce^(-3x)+e^(x)cosx+e^(x)sinx

Then for Y(p) I would use Axe^(-3x) correct?
Now, in reference to your question,
hils0005 said:
Y(c)=(r-3)(r^3+3r^2+r+3)
(r-3)(r+3)(r^2+1), r=3, -3, 1+i
Better :approve:. However,
hils0005 said:
y(c)=Ce^(3x)+Ce^(-3x)+e^(x)cosx+e^(x)sinx
You need to be careful with your constant coefficients here. The coefficients of the two exponentials need not be equal. Furthermore, you are missing the coefficients of the trigonometric terms.
hils0005 said:
Then for Y(p) I would use Axe^(-3x) correct?
Yes, since e-3x has already been found to be a solution, you cannot use e-3x as your particular solution. Axe-3x would be the usual alternative.
 
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dumb mistakes!
r=+/- 3, and 1+i
Y(c)=Ce^3x + Ce^-3x + Ce^(x)cosx + Ce^(x)sin(x)
Y(p) =Axe^-3x
y'(p) =Ae^-3x - 3Axe^-3x
y''(p) =-3Ae^-3x - 3Ae^-3x + 9Axe^-3x
y'''(p) =9Ae^-3x + 9Ae^-3x + 9Ae^-3x - 27Axe^-3x
y''''(p)=-27Ae^-3x -27Ae^-3x - 27Ae^-3x - 27Ae^-3x + 81Axe^-3x=
81AXe^-3x - 108Ae^-3x

substitute:
(81Axe^-3x - 108Ae^-3x)-8(9Axe^-3x - 6Ae^-3x) - 9(Axe^-3x)=7e^-3x

-60Ae^-3x=7e^-3x
-60A=7, A=-7/60

Answer:
y=C(1)e^3x + C(2)e^-3x + C(3)e^(x)cos(x) + C(4)e^(x)sinx - 7/60xe^-3x

does this look correct,
 
Last edited:
HallsofIvy
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dumb mistakes!
r=+/- 3, and 1+i
Back up! Your original equation was [itex]r^4- 8r^2- 9= (r^2- 9)(r^2+ 1)= 0[/itex]
so [itex]r^2= 9[/itex] and [itex]r^2= -1[/itex]. The roots of the first of those are [itex]\pm 3[/itex] but the roots of the second are not [itex]1\pm i[/itex]!
 
62
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the sqrt of -1= 1i correct-----ohhhhh wow that changes things:
e^(0)cos(x)+e^0sinx= C(3)cosx + C(4)sinx(x)
y=c(1)e^3x + c(2)e^-3x + c(3)cosx + c(4)sinx -7/60xe^-3x ??
 
Hootenanny
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Back up! Your original equation was [itex]r^4- 8r^2- 9= (r^2- 9)(r^2+ 1)= 0[/itex]
so [itex]r^2= 9[/itex] and [itex]r^2= -1[/itex]. The roots of the first of those are [itex]\pm 3[/itex] but the roots of the second are not [itex]1\pm i[/itex]!
:redface::shy::blushing:

That's really really really embarrassing.
 
Last edited:

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