4th Order Variation of Parameters

  • Thread starter VitaX
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  • #1
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Find the complementary solution of [itex]y^\left(4\right) + 2y'' + y = sint[/itex]

Homogeneous Form would be [itex]y^\left(4\right) + 2y'' + y = 0[/itex]

[itex]r^4 + 2r^2 + r = 0 \rightarrow r(r^3 + 2r + 1) = 0[/itex]

This is where I'm stuck. Once I find [itex]y_c(t)[/itex] I should be able to finish the problem, but I'm having trouble at this step. What would be the next step here?

The book's solution is [itex]y_c(t) = C_1 cost + C_2 sint + C_3 tcost + C_4 tsint[/itex] which would suggest complex numbers involved here.

Edit: Found my error, it was in the r equation.
 
Last edited:

Answers and Replies

  • #2
lurflurf
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(D4+2D2+1)=(D2+1)2
since sine is already a solution consider a particular solution of the form
At^2 sin t+Bt^2 cos t

edit:I forgot you wanted variation of parameters
just take
[itex]y(t) = C_1(t) cost + C_2(t) sint + C_3(t) tcost + C_4(t) tsint[/itex]
complex numbers are optional, you just need to be able to solve
(D2+1)y=0
to deal with quadratic terms that are irreducible over reals
 
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  • #3
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After taking the particular solution to be:

[itex]y_p(t) = U_1 (t)cost + U_2 (t)sint + U_3 (t)tcost + U_4 (t)tsint[/itex]

I then take the 4 derivatives and I ended up with the following 4 Equations (1st 3 are from the conditions that the U(t) derivatives add up to 0 while the last equation is derived from inputting the the derivatives into the original DE and cancelling out like terms):

[itex]U'_1 (t)cost + U'_2 (t)sint + U'_3 (t)tcost + U'_4 (t)tsint = 0[/itex]

[itex]-U'_1 (t)sint + U'_2 (t)cost + U'_3 (t)cost - U'_3 (t)tsint + U'_4 (t)sint + U'_4 (t)tcost = 0[/itex]

[itex]-U'_1 (t)cost - U'_2 (t)sint - 2U'_3 (t)sint - U'_3 (t)tscost + 2U'_4 (t)cost - U'_4 (t)tsint = 0[/itex]

[itex]U'_1 (t)sint - U'_2 (t)cost - 3U'_3 (t)cost + U'_3 (t)tsint - 3U'_4 (t)sint - U'_4 (t)tcost = sint[/itex]

But at this point, it is a complete monster. Can anyone possibly solve this mess?
 
  • #4
lurflurf
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It is a bit messy, but the differential equations part it settled and only algebra remains. It also helps that only one right hand side is nonzero. It is also possible to solve
(D2+1)y=f and then solve it again hence solving
(D2+1)2y=f
 
  • #5
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It is a bit messy, but the differential equations part it settled and only algebra remains. It also helps that only one right hand side is nonzero. It is also possible to solve
(D2+1)y=f and then solve it again hence solving
(D2+1)2y=f
Hmm, what exactly do you mean by those 2 equations?
 
  • #6
lurflurf
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(D2+1)y=f
y''+y=f
(D2+1)2y=f
y''''+2y''+y=f
the second can be reduced to the first
if
u''+u=f
and
y''+y=u
then
y''''+2y''+y=f

or just do all that algebra
 

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