3rd order differential eqn (wronskian)

[glitchy]

y''' + 25y' = csc(5x)

i got the y (complimentary) = C1 + C2cos5x + C3sin5x. I'm just having minor difficulties getting the y (particular).

 

[HallsofIvy]

Only minor difficulties? I don't see how you can expect anyone to help you if you won't say what you have done. What difficulties have you had? What method did you try?

Since csc(5x) is NOT one of the function that typically satisfy a "homogeneous linear equation with constant coefficients" (exponentials, polynomials, sine and cosine, and combinations of those) you can't "guess" a solution to use "undetermined coefficients". Did you try "variation of parameters"? That is, look for a solution of the form
y(x)= u(x)+ v(x)cos(5x)+ w(x) sin(5x) where u, v, w are unknown functions of x.
(there exist an infinite number of such functions that will satisfy the equation.)

(I presume you meant sin(5x) rather than cos(5x) twice!)

y'= u'+ v'cos(5x)- 5v(x) sin(5x)+ w' sin(5x)+ 5w(x) cos(5x)
Narrow the search to those functions such that
u'+ v'cos(5x)+ w' sin(5x)= 0 so that
y'= -5v(x) sin(5x)+ 5w(x) cos(5x) and
y"= -5v' sin(5x)- 25v(x)cos(5x)+ 5w' cos(5x)- 25 w(x)sin(5x)

Again narrow the search by requiring that
-5v' sin(5x)+ 5w' cos(5x)= 0 so that
y"= -25 v(x) cos(5x)-25w(x) sin(5x) and differentiate once more:

y'''= -25v' cos(5x)+ 125v sin(5x)- 25w' sin(5x)- 125 w cos(5x)

Putting those into the differential equation, because 1, sin(5x), and cos(5x) satisfy the homogenous equation, all terms involving only v and w will cancel leaving a single equation for v' and w'. That, together with the equations
u'+ v'cos(5x)+ w' sin(5x)= 0 and -5v' sin(5x)+ 5w' cos(5x)= 0 give you three equations you can solve for u', v', and w' and then integrate.

 

[glitchy]

so far i got
W = 125
W1 = 5csc5x
W2 = -5cos5xsin5x=-5(cos5x)^2 . tan5x
W3 =-5

therefor

U'1 = csc5x/25
U'2 = -cos5xsin5x/25
U'3 = -1/25



y=C1 + C2cos5x + C3sin5x + (1/5)Intan(5x/2) + (1/250)(cos5x)^3 - (1/25)xsin5x

i got that answer. can anyone verify that for me.