Filetype PDF | Posted on 30 Jan 2023 | 2 years ago
Authentication
digital resources
130x Filetype PDF File size 0.08 MB Source: www.math.utah.edu
4.6 Variation of Parameters 195
4.6 Variation of Parameters
The method of variation of parameters applies to solve
(1) a(x)y′′ + b(x)y′ + c(x)y = f(x):
Continuity of a, b, c and f is assumed, plus a(x) 6= 0. The method is
important because it solves the largest class of equations. Specifically
x2
included are functions f(x) like ln|x|, |x|, e .
Homogeneous Equation. The method of variation of parameters
uses facts about the homogeneous differential equation
(2) a(x)y′′ + b(x)y′ + c(x)y = 0:
The success depends upon writing the general solution of (2) as
(3) y = c y (x) +c y (x)
1 1 2 2
where y , y are known functions and c , c are arbitrary constants. If
1 2 1 2
a, b, c are constants, then the standard recipe for (2) finds y1, y2. It is
known that y1, y2 as reported by the recipe are independent.
Independence. Twosolutionsy1,y2of(2)arecalledindependentif
neither is a constant multiple of the other. The term dependent means
not independent, in which case either y1(x) = cy2(x) or y2(x) = cy1(x)
holds for all x, for some constant c. Independence can be tested through
the Wronskian of y1, y2, defined by
W(x)=y (x)y′(x)−y′(x)y (x):
1 2 1 2
Theorem 13 (Wronskian and Independence)
TheWronskianoftwosolutionssatisfiesa(x)W′+b(x)W = 0, whichimplies
Abel’s identity
−Rx(b(t)=a(t))dt
W(x)=W(x0)e x0 :
Two solutions of (2) are independent if and only if W(x) 6= 0.
The proof appears on page 183.
Theorem 14 (Variation of Parameters Formula)
Let a, b, c, f be continuous near x = x0 and a(x) 6= 0. Let y1, y2 be
two independent solutions of the homogeneous equation ay′′+by′+cy = 0
and let W(x) = y (x)y′(x) − y′(x)y (x). Then the non-homogeneous
1 2 1 2
differential equation
ay′′ + by′ + cy = f
has a particular solution
(4) yp(x) = y1(x)Z y2(x)(−f(x))dx+y2(x)Z y1(x)f(x)dx:
a(x)W(x) a(x)W(x)
196
The proof is delayed to page 183.
History of Variation of Parameters. The solution yp was dis-
covered by varying the constants c , c in the homogeneous solution (3),
1 2
assuming they depend on x. This results in formulas c (x) = R C F,
1 1
c (x) = R C F where F(x) = f(x)=a(x), C (t) = −y2(t),C (t) = y1(t);
2 2 1 W(t) 2 W(t)
see the historical details on page 183. Then
y = y (x)Z C F +y (x)Z C F Substitute in (3) for c , c .
1 1 2 2 1 2
=−y1(x)Z y2 F +y2(x)Z y1 F Use (??) for C1, C2.
W W
=Z (y2(x)y1(t)−y1(x)y2(t)) F(t) dt Collect on F=W.
W(t)
=Z y1(t)y2(x)−y1(x)y2(t)F(t)dt Expand W = y1y′ −y′y2.
y (t)y′(t) − y′(t)y (t) 2 1
1 2 1 2
Any one of the last three equivalent formulas is called a classical vari-
ation of parameters formula. The fraction in the last integrand is
called Cauchy’s kernel. We prefer the first, equivalent to equation (4),
for ease of use.
18 Example (Independence) Consider y′′ − y = 0. Show the two solutions
sinh(x) and cosh(x) are independent using Wronskians.
Solution: Let W(x) be the Wronskian of sinh(x) and cosh(x). The calculation
below shows W(x) = −1. By Theorem 10, the solutions are independent.
Background. The calculus definitions for hyperbolic functions are sinhx =
x −x x −x ′
(e −e )=2, coshx = (e +e )=2. Their derivatives are (sinhx) = coshx
′ ′ 1 x −x ′
and (coshx) = sinhx. For instance, (coshx) stands for 2(e + e ) , which
evaluates to 1(ex −e−x), or sinhx.
2
Wronskian detail. Let y1 = sinhx, y2 = coshx. Then
W=y(x)y′(x)−y′(x)y (x) Definition of Wronskian W.
1 2 1 2
=sinh(x)sinh(x)−cosh(x)cosh(x) Substitute for y , y′, y , y′.
1 1 2 2
1 x −x 2 1 x −x 2
= 4(e −e ) − 4(e +e ) Apply exponential definitions.
=−1 Expand and cancel terms.
19 Example (Wronskian) Given 2y′′ − xy′ + 3y = 0, verify that a solution
x2=4
pair y1, y2 has Wronskian W(x) = W(0)e .
Solution: Let a(x) = 2, b(x) = −x, c(x) = 3. The Wronskian is a solution
of W′ = −(b=a)W, hence W′ = xW=2. The solution is W = W(0)ex2=4, by
growth-decay theory.
4.6 Variation of Parameters 197
20 Example (Variation of Parameters) Solve y′′ +y = secx by variation of
parameters, verifying y = c cosx + c sinx+xsinx+cos(x)ln|cosx|.
1 2
Solution:
Homogeneous solution y . The recipe for constant equation y′′ + y = 0
h
is applied. The characteristic equation r2 + 1 = 0 has roots r = ±i and
yh = c1cosx+c2sinx.
Wronskian. Suitable independent solutions are y1 = cosx and y2 = sinx,
2 2
taken from the recipe. Then W(x) = cos x+sin x = 1.
Calculate yp. The variation of parameters formula (4) is applied. The inte-
gration proceeds near x = 0, because sec(x) is continuous near x = 0.
yp(x) = −y1(x)R y2(x)sec(x)dx+y2(x)R y1(x)secxdx 1
=−cosxR tan(x)dx+sinxR 1dx 2
=xsinx+cos(x)ln|cosx| 3
Details: 1 Use equation (4). 2 Substitute y1 = cosx, y2 = sinx. 3 Integral
tables applied. Integration constants set to zero.
′′ x
21 Example (Two Methods) Solve y −y = e by undetermined coefficients
and by variation of parameters. Explain any differences in the answers.
Solution: The general solution is reported to be y = yh +yp = c1ex +c2e−x+
xex=2. Details follow.
Homogeneoussolution. Thecharacteristicequationr2−1 = 0 for y′′−y = 0
has roots ±1. The homogeneous solution is yh = c1ex +c2e−x.
Undetermined Coefficients Summary. The basic trial solution method
gives initial trial solution y = d ex, because the RHS = ex has all derivatives
1
given by a linear combination of the independent function ex. The fixup rule
applies because the homogeneous solution contains duplicate term c ex. The
1
final trial solution is y = d xex. Substitution into y′′ − y = ex gives 2d ex +
1 1
d xex − d xex = ex. Cancel ex and equate coefficients of powers of x to find
1 1
d =1=2. Then y =xex=2.
1 p
Variation of Parameters Summary. Thehomogeneoussolutionyh = c1ex+
c e−x found above implies y = ex, y = e−x is a suitable independent pair of
2 1 2
solutions. Their Wronskian is W = −2
The variation of parameters formula (11) applies:
yp(x) = exZ −e−xexdx+e−xZ ex exdx:
−2 −2
Integration, followed by setting all constants of integrationto zero, gives yp(x) =
xex=2−ex=4.
Differences. The two methods give respectively yp = xex=2 and yp(x) =
xex=2−ex=4. The solutions yp = xex=2 and yp(x) = xex=2−ex=4 differ by the
homogeneous solution −xex=4. In both cases, the general solution is
y = c ex +c e−x + 1xex;
1 2 2
198
because terms of the homogeneous solution can be absorbed into the arbitrary
constants c1, c2.
Proof of Theorem 10: The function W(t) given by Abel’s identity is the
unique solution of the growth-decay equation W′ = −(b(x)=a(x))W; see page
3. It suffices then to show that W satisfies this differential equation. The
details:
′ ′ ′ ′
W =(y1y2−y1y2) Definition of Wronskian.
=y y′′ +y′y′ −y′′y −y′y′ Product rule; y′y′ cancels.
1 2 1 2 1 2 1 2 1 2
=y (−by′ −cy )=a−(−by′ −cy )y =a Both y , y satisfy (2).
1 2 2 1 1 2 1 2
=−b(y y′ −y′y )=a Cancel common cy y =a.
1 2 1 2 1 2
=−bW=a Verification completed.
The independence statement will be proved from the contrapositive: W(x) = 0
for all x if and only if y1, y2 are not independent. Technically, independence is
defined relative to the common domain of the graphs of y1, y2 and W. Hence-
forth, for all x means for all x in the common domain.
Let y1, y2 be two solutions of (2), not independent. By re-labelling as necessary,
y1(x) = cy2(x) holds for all x, for some constant c. Differentiation implies
y′(x) = cy′(x). Then the terms in W(x) cancel, giving W(x) = 0 for all x.
1 2
Conversely, let W(x) = 0 for all x. If y1 ≡ 0, then y1(x) = cy2(x) holds for
c = 0 and y , y are not independent. Otherwise, y (x ) 6= 0 for some x .
1 2 1 0 0
Define c = y (x )=y (x ). Then W(x ) = 0 implies y′(x ) = cy′(x ). Define
2 0 1 0 0 2 0 1 0
y = y −cy . By linearity, y is a solution of (2). Further, y(x ) = y′(x ) = 0.
2 1 0 0
Byuniqueness of initial value problems, y ≡ 0, that is, y2(x) = cy1(x) for all x,
showing y1, y2 are not independent.
Proof of Theorem 11: LetF(t) = f(t)=a(t), C (x) = −y (x)=W(x), C (x) =
1 2 2
y1(x)=W(x). Then yp as given in (4) can be differentiated twice using the
R ′
product rule and the fundamental theorem of calculus rule ( g) = g. Because
y C +y C =0andy′C +y′C =1,then y and its derivatives are given by
1 1 2 2 1 1 2 2 p
y (x) = y R C Fdx+y R C Fdx;
p 1 1 2 2
y′ (x) = y′ R C Fdx+y′ R C Fdx;
p 1 1 2 2
y′′(x) = y′′R C Fdx+y′′R C Fdx+F(x):
p 1 1 2 2
Let F =ay′′ +by′ +cy , F = ay′′ +by′ +cy . Then
1 1 1 1 2 2 2 2
ay′′ + by′ + cy = F Z C Fdx+F Z C Fdx+aF:
p p p 1 1 2 2
Because y1, y2 are solutions of the homogeneous differential equation, then
F =F =0. By definition, aF =f. Therefore,
1 2
ay′′ + by′ + cy = f:
p p p
The proof is complete.
Historical Details. The original variation ideas, attributed to Joseph Louis
Lagrange (1736-1813), involve substitution of y = c1(x)y1(x) +c2(x)y2(x) into
(1) plus imposing an extra condition on the unknowns c1, c2:
c′ y +c′y = 0:
1 1 2 2
no reviews yet
Please Login to review.
