Solving Tough Differential Equations with Variation of Parameters | [SOLVED]

[hils0005]

[SOLVED] Tough Diff EQ problem

Homework Statement

y^(4)-8y^(2)-9y=7e^(-3x)

y^(4)=fourth derivative, y^(2)=second deriv.

Homework Equations

Need to find complimentary solution and particular solution using variation of parameters

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)

 

[Hootenanny]

Are you sure about your homogenous solution? I suggest that you re-check your solutions of the indicial equation.

 

[hils0005]

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]

In reference to the PM, perhaps it would help if I draw your attention to the following lines

r^4-8r^2-9=0
(r^2-9)(r^2+1)=0

Good

r=9,-1

Not so, good. Notice the powers of r.

 

[Hootenanny]

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.

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,

Y(c)=(r-3)(r^3+3r^2+r+3)
(r-3)(r+3)(r^2+1), r=3, -3, 1+i

Better . However,

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.

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.

 

[hils0005]

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,

 

[HallsofIvy]

dumb mistakes!
r=+/- 3, and 1+i

Back up! Your original equation was $r^4- 8r^2- 9= (r^2- 9)(r^2+ 1)= 0$
so $r^2= 9$ and $r^2= -1$. The roots of the first of those are $\pm 3$ but the roots of the second are not $1\pm i$!

 

[hils0005]

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]

Back up! Your original equation was $r^4- 8r^2- 9= (r^2- 9)(r^2+ 1)= 0$
so $r^2= 9$ and $r^2= -1$. The roots of the first of those are $\pm 3$ but the roots of the second are not $1\pm i$!

:shy:

That's really really really embarrassing.