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Standard Letter Document Class for LATEX version 2e Leslie Lamport and Frank Mittelbach and Rainer Schopf July 1, 1998 Contents 1 Initial Code 2 1.1 Setting Paper Sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Choosing the type size . . . . . . . . . . . . . . . . . .

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• Publié par: osso0
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4.6 Variation of Parameters

4.6 Variation of Parameters

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Themethod of variation of parametersapplies to solve ′′ ′ (1)a(x)y+b(x)y+c(x)y=f(x). Continuity ofa,b,candfis assumed, plusa(x)6The method is= 0. important because it solves the largest class of equations.Specifically 2 x includedare functionsf(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 1 y 1 (x) +c 2 y 2 (x) wherey 1 ,y 2 areknown functionsandc 1 ,c 2 are arbitrary constants.If a,b,care constants, then the standardrecipefor (2) findsy 1 ,y 2 . Itis known thaty 1 ,y 2 as reported by the recipe areindependent.

Independence. Two solutionsy 1 ,y 2 of (2) are calledindependentif neither is a constant multiple of the other.The termdependentmeans not independent, in which case eithery 1 (x) =cy 2 (x) ory 2 (x) =cy 1 (x) holds for allx, for some constantc. Independencecan be tested through theWronskianofy 1 ,y 2 , defined by ′ ′ W(x) =y 1 (x)y(x) 2 −y 1 (x)y 2 (x). Theorem 13 (Wronskian and Independence) ′ The Wronskian of two solutions satisfiesa(x)W+b(x)W= 0, which implies Abel’s identity R x −(b(t)/a(t))dt x 0 W(x) =W(x 0 )e . Two solutions of (2) are independent if and only ifW(x)6= 0.

The proof appears on page 183. Theorem 14 (Variation of Parameters Formula) Leta,b,c,fbe continuous nearx=x 0 anda(x)6= 0. Lety 1 ,y 2 be ′′ ′ two independent solutions of the homogeneous equationay+by+cy= 0 ′ ′ W(x) =y 1 (x)y(x) (x)y and let 2 −y 1 2 (x)the nonhomogeneous. Then differential equation ′′ ′ ay+by+cy=f has a particular solution Z Z  y 2 (x)(−f(x))y 1 (x)f(x) (4)y p (x) =y 1 (x)dx+y 2 (x)dx . a(x)W(x)a(x)W(x)

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The proof is delayed to page 183.

History of Variation of Parameters. The solutiony p was dis covered by varying the constantsc 1 ,c 2 in the homogeneous solution (3), R assuming they depend onxresults in formulas. Thisc 1 (x) =C 1 F, R −y 2 (t)y 1 (t) c 2 (x) =C 2 FwhereF(x) =f(x)/a(x),C 1 (t,) =C 2 (t) =; W(t)W(t) see the historical details on page 183.Then Z Z y=y 1 (x)C 1 F+y 2 (x)C 2 FSubstitute in (3) forc 1 ,c 2 . Z Z F F =−y 1 (x)y 2 +y 2 (x)y 1 Use (??) forC 1 ,C 2 . W W Z F(t) = (y 2 (x)y 1 (t)−y 1 (x)y 2 (t))dtCollect onF/W. W(t) Z y 1 (t)y 2 (x)−y 1 (x)y 2 (t) ′ ′ =−y y. ′ ′ F(t)dtExpandW=y 1 y 1 2 2 y(t)y(t 1 2 )−y(t)y 2 (t) 1

Any one of the last three equivalent formulas is called aclassical vari ation of parameters formula. Thefraction in the last integrand is called Cauchy’s kernel.We prefer the first, equivalent to equation (4), for ease of use.

′′ 18 Example(Independence)Considery−y= 0. Showthe two solutions sinh(x)andcosh(x)are independent using Wronskians.

Solution: LetW(x) be the Wronskian of sinh(x) and cosh(xcalculation). The below showsW(x) =−Theorem 10, the solutions are independent.1. By Backgroundcalculus. Thedefinitionsfor hyperbolic functions are sinhx= x−x x−x′ (e−e)/2, coshx= (e+e)/derivatives are (sinh2. Theirx) =coshx ′ ′ 1 x−x′ and (coshxsinh) =x. Forinstance, (coshx) standsfor (e+ewhich) , 2 1 x−x evaluates to(e−e), or sinhx. 2 Wronskian detail. Lety 1 = sinhx,y 2 = coshx. Then ′ ′ (x) (x)y(x) W=y 1 (x)y 2 −y 1 2 Definition of WronskianW. ′ ′ nh(x)−cosh(x) cosh(x)Substitute fory,y, = sinh(x) si 1 1 y 2 ,y 2 . 1 x−x 2 1 x−x2 = (e−e)−(e+e)Apply exponential definitions. 4 4 =−1Expand and cancel terms.

′′ ′ 19 Example(Wronskian)Given2y−xy+ 3y= 0, verify that a solution 2 x /4 pairy 1 ,y 2 has WronskianW(x) =W(0)e.

Solution: Leta(x) = 2,b(x) =−x,c(x) = 3.The Wronskian is a solution 2 ′ ′x /4 ofW=−(b/a)W, henceW=xW/2. Thesolution isW=W(0)e, by growthdecay theory.

4.6 Variation of Parameters

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′′ 20 Example(Variation of Parameters)Solvey+y= secxby variation of parameters, verifyingy=c 1 cosx+c 2 sinx+xsinx+ cos(x) ln|cosx|.

Solution: ′′ Homogeneous solutiony h . Therecipefor constant equationy+y= 0 2 is applied.The characteristic equationr+ 1= 0 has rootsr=±iand y h =c 1 cosx+c 2 sinx. Wronskianindependent solutions are. Suitabley 1 = cosxandy 2 = sinx, 2 2 taken from therecipe. ThenW(x) = cosx+ sinx= 1. Calculatey p . TheThe intevariation of parameters formula (4) is applied. gration proceeds nearx= 0, because sec(x) is continuous nearx= 0. R R y p (x) =−y 1 (x)y 2 (x) sec(x)dx+y 2 (x)y 1 (x) secxdx1 R R =−cosxtan(x)dx+ sinx1dx2 =xsinx+ cos(x) ln|cosx|3

Details:1Use equation (4).2Substitutey 1 = cosx,y 2 = sinx.3Integral tables applied.Integration constants set to zero.

′′x 21 Example(Two Methods)Solvey−y=eby undetermined coefficients and by variation of parameters.Explain any differences in the answers. x−x Solution: The general solution is reported to bey=y h +y p =c 1 e+c 2 e+ x xe /follow.2. Details 2′′ Homogeneous solution. Thecharacteristic equationr−1 = 0 fory−y= 0 x−x has roots±1. The homogeneoussolution isy h =c 1 e+c 2 e. Undetermined Coefficients Summarybasic trial solution method. The x x gives initial trial solutiony=d 1 e, because the RHS =ehas all derivatives x given by a linear combination of the independent functionefixup rule. The x applies because the homogeneous solution contains duplicate termc 1 e. The x′′x x final trial solution isy=d 1 xeinto. Substitutiony−y=egives 2d 1 e+ x xx x d 1 xe−d 1 xe=e. Canceleand equate coefficients of powers ofxto find x d 1 = 1/2. Theny p =xe /2. x Variation of Parameters Summary. Thehomogeneous solutiony h =c 1 e+ −x x−x c 2 efound above impliesy 1 =e,y 2 =eis a suitable independent pair of solutions. Their WronskianisW=−2 The variation of parameters formula (11) applies: Z Z −x x −e e x x−x x y p (x) =e edx+dx.e e −2−2 Integration, followed by setting all constants of integration to zero, givesy p (x) = x x xe /2−e /4. x Differences. Thetwo methods give respectivelyy p =xe /2 andy p (x) = x xx xx xe /2−e /4. Thesolutionsy p =xe /2 andy p (x) =xe /2−e /4 differ by the x homogeneous solution−xe /the general solution is4. In both cases, 1 x−x x y=c 1 e+c 2 e+xe , 2

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because terms of the homogeneous solution can be absorbed into the arbitrary constantsc 1 ,c 2 . Proof of Theorem 10: The functionW(t) given by Abel’s identity is the ′ unique solution of the growthdecay equationW=−(b(x)/a(x))W; see page 3. Itsuffices then to show thatWsatisfies this differential equation.The details: ′ ′ ′′ −y W= (y 1 y 2 1 y 2 )Definition of Wronskian. ′′ ′′ ′′′ ′′ ′ =y y+y y−yduct rule;y ycancels. 1 2 1 21 y 2 −y yPro 1 2 1 2 ′ ′ −cy 1 )y 2 /aBothy 1 ,y 2 satisfy (2). =y 1 (−by−cy 2 )/a−(−by 1 2 ′ ′ y−y y)/aCan =−b(y 1 21 2 cel commoncy 1 y 2 /a. =−bW/aVerification completed. The independence statement will be proved from the contrapositive:W(x) = 0 for allxif and only ify 1 ,y 2 are not independent.Technically, independence is defined relative to the common domain of the graphs ofy 1 ,y 2 andW. Hence forth,for allxmeans for allxin the common domain. Lety 1 ,y 2 By relabelling as necessary,be two solutions of (2), not independent. y 1 (x) =cy 2 (x) holds for allx, for some constantc. Differentiationimplies ′ ′ y(x) =cy(x). Thenthe terms inW(x) cancel, givingW(x) = 0 for allx. 1 2 Conversely, letW(x) = 0 for allx. Ify 1 ≡0, theny 1 (x) =cy 2 (x) holds for c= 0 andy 1 ,y 2 are not independent.Otherwise,y 1 (x 0 )6= 0 for somex 0 . ′ ′ Definec=y(x)/y(x). esy( 2 01 0 ThenW(x 0 ) = 0 impli 1 (x 0 ). Define 2 x 0 ) =cy ′ y=y 2 −cy 1 linearity,. ByyFurther,is a solution of (2).y(x 0 ) =y(x 0 ) = 0. By uniqueness of initial value problems,y≡0, that is,y 2 (x) =cy 1 (x) for allx, showingy 1 ,y 2 are not independent. Proof of Theorem 11: LetF(t) =f(t)/a(t),C 1 (x) =−y 2 (x)/W(x),C 2 (x) = y 1 (x)/W(x). Theny p 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= 0+y C= 1, thenyand its derivatives are given by y 1 C 21 2 andy 1 C 1 22p R R y p (x) =y 1 C 1 F dx+y 2 C 2 F dx, R R ′ ′′ y(x) =y p1 C 1 F dx+y C 2 F dx, 2 R R ′′ ′′′′ y CF dx+F(x). y(x) =y C 1 F dx+ 2 2 p1 ′′ ′′′ ′ +by+cy,F=ay+by+cy. Then LetF 1 =ay 2 2 2 1 1 1 2 Z Z ′′ ′ ay+by+cy p =F 1 C 1 F dx+F 2 C 2 F dx+aF. p p Becausey 1 ,y 2 are solutions of the homogeneous differential equation, then F 1 =F 2 = 0.By definition,aF=f. Therefore, ′′ ′ + ay p by+cy p =f. p The proof is complete. Historical Details. The original variation ideas, attributed to Joseph Louis Lagrange (17361813), involve substitution ofy=c 1 (x)y 1 (x) +c 2 (x)y 2 (x) into (1) plus imposing an extra condition on the unknownsc 1 ,c 2 : ′ ′ y+c c 1 1 2 y 2 = 0.

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′ ′′ ′′ The product rule givesy=c y 1 +c 1 y+c y 2 +c 2 y, which then reduces to 1 12 2 ′ ′ ′ pressiony=c y. Substitutioninto (1) gives the twotermed ex 1 1 +c 2 y 2 ′ ′′′ ′′ ′′′ ′ a(c y+c 1 y+c y+c 2 y) +b(c 1 y+c 2 y) +c(c 1 y 1 +c 2 y 2 ) =f 1 11 22 21 2 which upon collection of terms becomes ′′ ′′′ ′′ ′′ ′ c(a+ 1 2 (ay+by 2 +cy 2 ) +ay c+ay c=f. 1 y+by 1 cy 1 ) +c 2 11 22 The first two groups of terms vanish becausey 1 ,y 2 are solutions of the homo ′ ′′ ′ geneous equation, leaving justay c+ay c=fare now two equations. There 1 12 2 ′ ′ and two unknownsX=c,Y=c: 1 2 ′ ′ ay X+ay Y=f, 1 2 y 1 X+y 2 Y= 0. Solving by elimination, −y 2 f y 1 f X=, Y=. aW aW Thenc 1 is the integral ofXandc 2 is the integral ofY, which completes the historical account of the relations Z Z −y2(x)f(x)y 1 (x)f(x) c 1 (x) =dx, c 2 (x) =dx. a(x)W(x)a(x)W(x)

Exercises 4.6 ′′ ′ Independence . Findsolutionsy 1 ,y 2 12.y+ 16y+ 4y= 0 of the given homogeneous differential 2′′ 13.x y+y= 0 equation which are independent by the Wronskian test, page 180. 2′′ 14.x y+ 4y= 0 ′′ 1.y−y= 0 2′′ ′ 15.x y+ 2xy+y= 0 ′′ 2.y−4y= 0 2′′ ′ 16.x y+ 8xy+ 4y= 0 ′′ 3.y+y= 0 ′′ Wronskian the Wronskian,. Compute 4.y+ 4y= 0 up a constant multiple, without solv ′′ ing the differential equation. 5.4y= 0 ′′ ′ 17.y+y−xy= 0 ′′ 6.y= 0 ′′ ′ ′′ ′ 18.y−y+xy= 0 7.4y+y= 0 ′′ ′ ′′ ′ 8.y+y= 0 19.2y+y+ sin(x)y= 0 ′′ ′′′ ′ 9.y+y+y= 020.4y−y+ cos(x)y= 0 ′′ ′ 2′′ ′ 10.y−y+y= 0 21.x y+xy−y= 0 ′′ ′ 2′′ ′ 11.y+ 8y+ 2y= 0 22.x y−2xy+y= 0