How can I use variation of parameters to solve this differential equation?

[Damascus Road]

Hey all,

this is a little confusing, because the "variation of parameters" that I have been taught in class is different then what I find in most texts...

I have y''' + y' = tan(x)

Most textbooks use the wronskian and work from there,
what I was taught to do is set it up as the characteristic eqn, and then factoring it I get solutions

r = 0, -i , + i

(Side question, may I set up my solution as:

y= C1 + C2sin(x) + C3cos(x) + C4 e^ix + C5 e^ - ix ?

or must it be something like...

y= C1 +C2 e^ix + C3 e^ - ix + C4 xe^ix + C5 xe^ - ix ? )

Anyways,

then when we take the derivatives, we end up with a system of equations, where the sum of each term with a derivative of a constant = 0,
and the last expression = tan x

But solving these is difficult...

HELP!

 

[djeitnstine]

First of all e^{ix} will simplify to sin and cos in each case to make life easier. And you should show us what you have done so far that we may help you. My way of doing it may be slightly different from yours.

 

[Damascus Road]

Ok,
well simplifying I get these 4 expressions::

C1' cos(x) + C2' sin(x) + C3' xcos(x) + C4' xsin(x) = 0

-C1' sin(x) + C2' cos(x) + C3' xcos(x) - C4' xsin(x) = 0

-C1' cos(x) - C2' sin(x) - C3' xcos(x) - C4' xsin(x) = tan(x)

Now I need to solve this system, and integrate for the constants...

 

[HallsofIvy]

I have no idea what you are doing! Yes, the characteristic equation is r³+r= 0 and has roots 0, i, and -i. That tells you that the general solution to the associated homogeneous equation would be y= C₀+ C₁ cos(x)+ C₂ sin(x). There is no reason for complex exponentials (those would reduce to the cosine and sine functions) nor for multiplying by x.

Now, to use "variation of parameters", look for a solution to the entire equation of the form y= u(x)+ v(x)cos(x)+ w(x)sin(x). y'= u'+ v' cos(x)- v sin(x)+ w' sin(x)+ w cos(x). There are, in fact, an infinite number of choices for u, v, and w and we can simplify by narrowing the search to only those that satisfy u'+ v' cos(x)+ w' sin(x)= 0. That leaves y'= -v sin(x)+ w cos(x) so now y"= -v' sin(x)- v cos(x)+ w' cos(x)- w sin(x). Again, we narrow to those u, v, w satisfying -v' sin(x)+ w' cos(x)= 0 and have y"= -v cos(x)- w sin(x). Differentiating one more time y"'= -v' cos(x)+ v sin(x)- w'sin(x)- w cos(x). Putting that and y'= -vsin(x)+ w cos(x) into the eqation, v' cos(x)+ v sin(x)- w' sin(x)- w cos(x)- vsin(x)+ wcos(x)= v' cos(x)- w'sin(x)= tan(x). We now have three equations:

u'+ v' cos(x)+ w' sin(x)= 0
-v' sin(x)+ w' cos(x)= 0
v' cos(x)- w' sin(x)= tan(x)
To solve for u', v' and w'.