Section 4.6 : Variation of Parameters
A method for solving nonhomogeneous linear differential equations that is more general than the method of undetermined coefficients.
nth-Order Linear Nonhomogeneous ODE in Standard Form:
y(n) + pn−1(x) y(n−1) + · · · + p1(x) y0 + p0(x) y = g(x) (1)
Method of undetermined coefficients can be used to find a particular solution yp to Eq. (1) when:
• The coefficients are all constants, i.e., pi(x) = ai, i = 0, ..., n − 1, where ai are constants.
• The right-hand side function g(x) is:
– a constant
– a polynomial function
– an exponential function, eαx
– a sine or cosine function, sin(βx), cos(βx) – a linear combination of these functions – a product of these functions
– a linear combination of a product of these functions That is,
c, xk, xkeαx, xkeαx cos(βx), xkeαx sin(βx)
or linear combinations thereof, depending on the roots of the auxiliary
Method of variation of parameters can be used to find a particular solution yp to Eq. (1) when:
• The coefficients are functions of x.
• The right-hand side function g(x) is any integrable function.
Notes:
1. Variation of parameters is a more general method then the method of undetermined coefficients, but it is less convenient in elementary appli- cations.
2. Unlike method of undetermined coefficients, variation of parameters will always yield a particular solution to the nonhomogeneous linear ODE, as long as the associated homogeneous ODE can be solved.
3. In both methods, the general solution yc of the associated homogeneous ODE must be known.
Consider Second-Order ODE in Standard Form:
y00 + P(x) y0 + Q(x) y = g(x) (2) (D2 + P(x) D + Q(x)) y = g(x) (2) Assume: The complementary function yc is known. That is, we know the general solution
yc(x) = c1 y1(x) + c2 y2(x) (3) to the associated homogeneous ODE, where y1 and y2 form the fundamental set of solutions.
Basic Idea of Method of Variation of Parameters:
Seek a particular solution to Eq. (2) of the form
yp(x) = u1(x) y1(x) + u2(x) y2(x) (4)
That is, allow the constant coefficients in Eq. (3) to vary.
We can find u1 and u2 as follows:
(A) Substitute yp given in Eq. (4) into ODE (2). This will yield a system of 2 equations in 2 unknowns, u01 and u02.
(B) Solve this system for u01 and u02.
(C) Integrate u01 and u02 to find u1 and u2, which can then be used in Eq. (4) to give exact form of yp.
(A) Substitute y = yp given in Eq. (4) into ODE (2) and derive a system of 2 equations in 2 unknowns, u01 and u02.
yp = u1 y1 + u2 y2
y0p = [u1 y01 + u01 y1] + [u2 y02 + u02 y2]
y00p = [u1 y001 + 2u01 y01 + u001 y1]+ [u2 y002 + 2u02 y02 + u002 y2]
Substitution into ODE (2) and rearranging terms yields:
g(x) = y00p + P(x) y0p + Q(x) yp
= u1
hy001 + Py01 + Qy1i + u2
hy002 + Py02 + Qy2
i + h
y1u001 + y01u01i + hy2u002 + y02 u02i + P h
y1u01 + y2u02i + y01u01 + y02u02
= d
dx
hy1u01i + d dx
hy2u02i + P hy1u01 + y2u02i + y01u01 + y02u02
= d
dx
hy1u01 + y2u02i + P hy1u01 + y2u02i + y01u01 + y02u02 (5)
We need two equations to determine u1 and u2.
Assume: u1 and u2 are such that: y1u01 + y2u02 = 0 Then Eq. (5) reduces to: y01u01 + y02u02 = g(x)
That is, Eq. (5) is satisfied when:
y1u01 + y2u02 = 0
y01u01 + y02u02 = g(x) (6)
This is a system of 2 equations in 2 unknowns, u01 and u02.
Matrix form:
" y1 y2 y01 y02
# " u01 u02
#
=
"
0 g(x)
#
(7)
(B) Solve this system for u01 and u02.
We can solve this system using
• Row reduction (i.e., Gaussian Elimination)
• Solving first equation for one unknown and substituting this into the second equation
• Cramer’s rule
In any case, the solution is:
u01 = −y2 g(x)
y1 y02 − y01 y2 and u02 = y1 g(x)
y1 y02 − y01 y2 (8)
Cramer’s Rule:
Expresses the solution to an n × n system of equations in terms of determinants.
Matrix form of system:
y1 y2 y01 y02
u01 u02
=
0 g(x)
(7)
Solution:
u01 = W1
W = −y2 g(x)
W and u02 = W2
W = y1 g(x)
W (9)
where
W =
y1 y2 y01 y02
= y1 y02 − y01 y2 (10)
W1 =
0 y2 g(x) y02
= −y2 g(x) (11)
W2 =
y1 0 y01 g(x)
= y1 g(x) (12)
Note: The determinant of the coefficient matrix is the Wronskian of y1 and y2.
Since y1 and y2 are linearly independent solutions to the associated homo- geneous ODE on some interval I, it follows that
W = W(y1(x), y2(x)) , 0 for all x in I.
(C) Integrate u01 and u02 to find u1 and u2. Then set
yp = u1 y1 + u2 y2 (4) to get exact form of yp.
Notes:
1. The ODE must be in standard form to apply this method.
2. In practice, the derivation of the system of equations in (A) is not per- formed every time an ODE is solved via variation of parameters, since it is too long.
Instead, the form of the system in Eq. (6) or Eq. (7) should be known.
For solving the system, it would be helpful to known how to apply Cramer’s rule, as given in Eqs. (9)–(12).
Example: Solve y00 − 3y0 + 2y = 1 1 + e−x
That is, find the general solution y = yc + yp.
Example:
Higher-Order Equations:
Linear nth-order ODE in standard form:
y(n) + pn−1(x) y(n−1) + · · · + p1(x) y0 + p0(x) y = g(x) (13)
Let y1, ..., yn be a fundamental set of solutions to the associated homoge- neous ODE, so that
yc = c1 y1 + c2 y2 + · · · + cn yn is the complementary function for Eq. (13).
Then a particular solution to Eq. (13) is:
yp = u1(x) y1(x) + u2(x) y2(x) + · · · + un(x) yn(x) (14)
where u0k, k = 1, ..., n, are determined by solving the n × n system of equa- tions:
y1u01 + y2u02 + · · · + ynu0n = 0 y01u01 + y02u02 + · · · + y0nu0n = 0
y001 u01 + y002 u02 + · · · + y00n u0n = 0
... ... ...
y(n−1)1 u01 + y(n−1)2 u02 + · · · + y(n−1)n u0n = g(x) (15)
Matrix form:
y1 y2 · · · yn y01 y02 · · · y0n y001 y002 · · · y00n
... ... ...
y(n−1)1 y(n−1)2 · · · y(n−1)n
u01 u02 ...
u0n
=
0 0 ...
g(x)
(16)
By Cramer’s rule, the solution is:
u0k = Wk
W , k = 1, ..., n (17)
where
• W is the determinant of the coefficient matrix in Eq. (16) (here: W is the Wronskian of y1, ..., yn);
• Wk is the determinant obtained by replacing the column k of the Wronskian by the right-hand side column vector in Eq. (16).
