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# Differential Equations : Variation of Parameters

Use variation of parameters to solve the differential equation y'''+4y'=cot2t.

#### Solution Preview

First, a trick:

reduce the order of the differential equation by substituting w(t) = y'(t)

Then y''' = w'', and now we can use the general method of solution of variation of parameters for a second-order equation (MUCH easier than playing with a third!) We solve for w and then backtrack to get y.

(Dirty trick, right? Remember these things when they come up -- they help a lot in odd situations.)

Given y''' + 4y' = cot2t

Then w'' + 4w = cot2t

Now we need two independent solutions to the related homogeneous equation
w'' + 4w = 0

Well, this is a well-known problem. We set up the quadratic
p^2 + 4 = 0
solve - p^2 = -4
solutions p = 2i, -2i

Well, The solutions are e^2it and e^-2it
remember e^(a + bi) = (e^a)(cosb + i*sinb)

So the solutions to the homogeneous equation are
w1 = cos2t + i*sin2t and w2 = cos(-2t) + i*sin(-2t)

(sorry this board does not make subscripts easy -- of course that's the usual w sub 1 and later the usual constant c sub 1 etc)

I tend to check everything:
w1' = -2sin2t + 2i*cost2t
w1'' = -4cos2t - 4i*sin2t
and we see that w1'' + 4w1 = 0, correct, similarly for w2, so OK on this part

Now the general solution of the homogeneous equation is
w = c1*w1 + c2 *w2

And now we need the thitd part, a particular solution to the original reduced equation.
staying with w's and back to y later.

I am taking variation of parameters from the classic DE text by Boyce and DiPrima.
Here's another hint for you -- if you're going to do this stuff, collect a library of classic texts. I may not remember every single thing after a decade or two, but I can go and get it back.

First we assume that the general solution is of the form

w = u1 * w1 + u2 * w2

where u1 and u2 are some functions of t
In order to make this soluble, we require u1'*w1 + u2'*w2 = 0

So w' = u1*w1' + u2*w2' + u1'*w1 + u2'*w2 , using the zero above
w' = u1*w1' + u2*w2'

Using our w1 and w2 above

w = u1 * w1 + u2 * w2 = u1cos2t + u1*i*sin2t + u2 cos(-2t) + u2*i*sin(-2t)

w' = u1*w1' + u2*w2' = -2u1sin2t + 2u1*i*cost2t + 2u2sin(-2t) - 2u2*i*cos(-2t)

Working out from this, awful product rule,
w'' = -4u1cos2t -4u1*i*sin2t -4u1cos(-2t) -4u1*i*sin(-2t) -2u1'sin2t + 2u1'*i*cost2t + 2u2'sin(-2t) - 2u2'*i*cos(-2t)

Substituting into the reduced equation

w'' + 4w = cot2t

-4u1cos2t -4u1*i*sin2t -4u2cos(-2t) -4u2*i*sin(-2t) -2u1'sin2t + 2u1'*i*cost2t + 2u2'sin(-2t) - 2u2'*i*cos(-2t) + 4(u1cos2t + u1*i*sin2t + u2 cos(-2t) + u2*i*sin(-2t) ) = ...

#### Solution Summary

Variation of parameters is used to solve an ODE. The solution is detailed and well presented.

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