Question

Variation of parameters of 4th order differential equation

Answer 1

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Answer 2

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.

Let 

        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' 

             |             |                  |

        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 

            |            |            |

        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.  

Example

Solve 

        y''' + y'  =  sec t

 

Solution

We first find the homogeneous solution.  The characteristic equation is 

        r3 + r  =  0

        r(r2 + 1)  =  0

        r  =  0,        r  =  i,         r  =  -i

We conclude 

        yh  =  c1 + c2 cos t + c3 sin t

We have

        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

        

So that 

        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 ti

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