Variation of Parameters for Higher Order Differential Equations
For second order nonhomogeneous differential equations, we saw that if the function g(x) does not generate a UC-Set, then we must use the method of variation of parameters. To review this method click here. It turns out that the method of variation of parameters naturally generalizes to higher order differential equations.
L(y) = y(n) + p1y(n-1) + ... + pn-1y' + pny = g(t)
and suppose that the general solution to L(y) = 0 is given by
yh = c1y1 + c2y2 + ... + cnyn
As with variation of parameters for second order equations, we assume that
yp = u1y1 + u2y2 + ... + unyn
where the ui are functions of t. Following the prior discussion, we see that we need only one yp but have labeled n functions of t. We therefore may impose n - 1 conditions on the u's. The conditions that we impose are
u1y1' + u2y2' + ... + unyn'
u1'y1' + u2'y2' + ... + un'yn'
u1''y1' + u2''y2' + ... + un''yn'
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u1(n-1) y1' + u2(n-1) y2' + ... + un(n-1) yn'
Now take derivatives to get
yp' = u1y1' + u1'y1 + u2y2' +u2'y2 + ... + unyn' + un'yn = u1'y1 + u2'y2 + ... + un'yn
yp'' = u1'y1' + u1''y1 + u2'y2' +u2''y2 + ... + un'yn' + un''yn = u1''y1 + u2''y2 + ... + un''yn
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yp(n-1) = u1(n-1)y1 + u2(n-1)y2 + ... + un(n-1)yn
Plugging in to the original differential equation and noting that each of the y's are solution to the differential equation, we get
y1(n-1)u1' + y2(n-1)u2' + ... + yn(n-1)un' = g(t)
Notice that this gives a system of n equations and n unknowns. We can write this as
Since the Wronskian is never zero, we can take its inverse. Taking the inverse of matrix where the entries are functions is not easy. Fortunately, we only need the last column of the inverse, which can be found with much less pain using the adjoint formula.
y''' + y' = sec t
We first find the homogeneous solution. The characteristic equation is
r3 + r = 0
r(r2 + 1) = 0
r = 0, r = i, r = -i
yh = c1 + c2 cos t + c3 sin t
yp = v1 + v2 cos t + v3 sin t
and the Wronskian matrix is
The Wronskian is
|W| = 1(sin2 t + cos2 t) - cos t (0 - 0) + sin t (0 - 0) = 1
The matrix equation takes the following form
u1' = sec t u2' = -1 u3' = -tan t
Integrating, we get
u1' = ln|sec t + tan t| u2' = -t u3' = ln|cos t|
The final solution is
y = c1 + c2 cos t + c3 sin t + ln|sec t + tan t| -t cos t + (sin t) ln|cos t|