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Standard Letter Document Class for LATEX version 2e

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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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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=c1y1(x) +c2y2(x) wherey1,y2areknown functionsandc1,c2are arbitrary constants.If a,b,care constants, then the standardrecipefor (2) findsy1,y2. Itis known thaty1,y2as reported by the recipe areindependent.
Independence.Two solutionsy1,y2of (2) are calledindependentif neither is a constant multiple of the other.The termdependentmeans not independent, in which case eithery1(x) =cy2(x) ory2(x) =cy1(x) holds for allx, for some constantc. Independencecan be tested through theWronskianofy1,y2, defined by ′ ′ W(x) =y1(x)y(x) 2y1(x)y2(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(x0)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=x0anda(x)6= 0. Lety1,y2be ′′ ′ two independent solutions of the homogeneous equationay+by+cy= 0 ′ ′ W(x) =y1(x)y(x) (x)y and let2y1 2(x)the nonhomogeneous. Then differential equation ′′ ′ ay+by+cy=f has a particular solution Z Z  y2(x)(f(x))y1(x)f(x) (4)yp(x) =y1(x)dx+y2(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 solutionypwas dis covered by varying the constantsc1,c2in the homogeneous solution (3), R assuming they depend onxresults in formulas. Thisc1(x) =C1F, R y2(t)y1(t) c2(x) =C2FwhereF(x) =f(x)/a(x),C1(t,) =C2(t) =; W(t)W(t) see the historical details on page 183.Then Z Z y=y1(x)C1F+y2(x)C2FSubstitute in (3) forc1,c2. Z Z F F =y1(x)y2+y2(x)y1Use (??) forC1,C2. W W Z F(t) = (y2(x)y1(t)y1(x)y2(t))dtCollect onF/W. W(t) Z y1(t)y2(x)y1(x)y2(t) ′ ′ =y y. ′ ′F(t)dtExpandW=y1y1 2 2 y(t)y(t 1 2)y(t)y2(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)Consideryy= 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= xx xx(ee)/2, coshx= (e+e)/derivatives are (sinh2. Theirx) =coshx ′ ′1xxand (coshxsinh) =x. Forinstance, (coshx) standsfor (e+ewhich) , 2 1xx evaluates to(ee), or sinhx. 2 Wronskian detail. Lety1= sinhx,y2= coshx. Then ′ ′ (x) (x)y(x) W=y1(x)y2y1 2Definition of WronskianW. ′ ′ nh(x)cosh(x) cosh(x)Substitute fory,y, = sinh(x) si1 1y2,y2. 1xx2 1xx2 = (ee)(e+e)Apply exponential definitions. 4 4 =1Expand and cancel terms.
′′ ′ 19 Example(Wronskian)Given2yxy+ 3y= 0, verify that a solution 2 x /4 pairy1,y2has 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=c1cosx+c2sinx+xsinx+ cos(x) ln|cosx|.
Solution: ′′ Homogeneous solutionyh. Therecipefor constant equationy+y= 0 2 is applied.The characteristic equationr+ 1= 0 has rootsr=±iand yh=c1cosx+c2sinx. Wronskianindependent solutions are. Suitabley1= cosxandy2= sinx, 2 2 taken from therecipe. ThenW(x) = cosx+ sinx= 1. Calculateyp. TheThe intevariation of parameters formula (4) is applied. gration proceeds nearx= 0, because sec(x) is continuous nearx= 0. R R yp(x) =y1(x)y2(x) sec(x)dx+y2(x)y1(x) secxdx1 R R =cosxtan(x)dx+ sinx1dx2 =xsinx+ cos(x) ln|cosx|3
Details:1Use equation (4).2Substitutey1= cosx,y2= sinx.3Integral tables applied.Integration constants set to zero.
′′x 21 Example(Two Methods)Solveyy=eby undetermined coefficients and by variation of parameters.Explain any differences in the answers. xx Solution:The general solution is reported to bey=yh+yp=c1e+c2e+ x xe /follow.2. Details 2′′ Homogeneous solution. Thecharacteristic equationr1 = 0 foryy= 0 xx has roots±1. The homogeneoussolution isyh=c1e+c2e. Undetermined Coefficients Summarybasic trial solution method. The x x gives initial trial solutiony=d1e, 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 termc1e. The x′′x x final trial solution isy=d1xeinto. Substitutionyy=egives 2d1e+ x xx x d1xed1xe=e. Canceleand equate coefficients of powers ofxto find x d1= 1/2. Thenyp=xe /2. x Variation of Parameters Summary. Thehomogeneous solutionyh=c1e+ x xx c2efound above impliesy1=e,y2=eis a suitable independent pair of solutions. Their WronskianisW=2 The variation of parameters formula (11) applies: Z Z x x e e x xx x yp(x) =e edx+dx.e e 22 Integration, followed by setting all constants of integration to zero, givesyp(x) = x x xe /2e /4. x Differences. Thetwo methods give respectivelyyp=xe /2 andyp(x) = x xx xx xe /2e /4. Thesolutionsyp=xe /2 andyp(x) =xe /2e /4 differ by the x homogeneous solutionxe /the general solution is4. In both cases, 1 xx x y=c1e+c2e+xe , 2
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because terms of the homogeneous solution can be absorbed into the arbitrary constantsc1,c2. 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= (y1y2 1y2)Definition of Wronskian. ′′ ′′ ′′′ ′′ ′ =y y+y yyduct rule;y ycancels. 1 21 21y2y yPro1 2 1 2 ′ ′ cy1)y2/aBothy1,y2satisfy (2). =y1(bycy2)/a(by1 2 ′ ′ yy y)/aCan =b(y1 21 2cel commoncy1y2/a. =bW/aVerification completed. The independence statement will be proved from the contrapositive:W(x) = 0 for allxif and only ify1,y2are not independent.Technically, independence is defined relative to the common domain of the graphs ofy1,y2andW. Hence forth,for allxmeans for allxin the common domain. Lety1,y2By relabelling as necessary,be two solutions of (2), not independent. y1(x) =cy2(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. Ify10, theny1(x) =cy2(x) holds for c= 0 andy1,y2are not independent.Otherwise,y1(x0)6= 0 for somex0. ′ ′ Definec=y(x)/y(x). esy( 2 01 0ThenW(x0) = 0 impli1(x0). Define 2x0) =cy y=y2cy1linearity,. ByyFurther,is a solution of (2).y(x0) =y(x0) = 0. By uniqueness of initial value problems,y0, that is,y2(x) =cy1(x) for allx, showingy1,y2are not independent. Proof of Theorem 11:LetF(t) =f(t)/a(t),C1(x) =y2(x)/W(x),C2(x) = y1(x)/W(x). Thenypas 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 y1C21 2andy1C1 22p R R yp(x) =y1C1F dx+y2C2F dx, R R ′ ′y(x) =y p1C1F dx+y C2F dx, 2 R R ′′ ′′′′ y CF dx+F(x). y(x) =y C1F dx+2 2 p1 ′′ ′′′ ′ +by+cy,F=ay+by+cy. Then LetF1=ay2 2 21 11 2 Z Z ′′ ′ ay+by+cyp=F1C1F dx+F2C2F dx+aF. p p Becausey1,y2are solutions of the homogeneous differential equation, then F1=F2= 0.By definition,aF=f. Therefore, ′′ ′ + aypby+cyp=f. p The proof is complete. Historical Details.The original variation ideas, attributed to Joseph Louis Lagrange (17361813), involve substitution ofy=c1(x)y1(x) +c2(x)y2(x) into (1) plus imposing an extra condition on the unknownsc1,c2: ′ ′ y+c c1 12y2= 0.
4.6 Variation of Parameters
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′ ′′ ′The product rule givesy=c y1+c1y+c y2+c2y, which then reduces to 1 12 2 ′ ′ ′ pressiony=c y. Substitutioninto (1) gives the twotermed ex1 1+c2y2 ′ ′′′ ′′ ′′′ ′ a(c y+c1y+c y+c2y) +b(c1y+c2y) +c(c1y1+c2y2) =f 1 11 22 21 2 which upon collection of terms becomes ′′ ′′′ ′′ ′′ ′ c(a+ 1 2(ay+by2+cy2) +ay c+ay c=f. 1y+by1cy1) +c 2 11 22 The first two groups of terms vanish becausey1,y2are 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 y1X+y2Y= 0. Solving by elimination, y2f y1f X=, Y=. aW aW Thenc1is the integral ofXandc2is the integral ofY, which completes the historical account of the relations Z Z y2(x)f(x)y1(x)f(x) c1(x) =dx, c2(x) =dx. a(x)W(x)a(x)W(x)
Exercises 4.6 ′′ ′ Independence. Findsolutionsy1,y212.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.yy= 0 2′′ ′ 15.x y+ 2xy+y= 0 ′′ 2.y4y= 0 2′′ ′ 16.x y+ 8xy+ 4y= 0 ′′ 3.y+y= 0 ′′Wronskianthe Wronskian,. Compute 4.y+ 4y= 0 up a constant multiple, without solv ′′ing the differential equation. 5.4y= 0 ′′ ′ 17.y+yxy= 0 ′′ 6.y= 0 ′′ ′ ′′ ′ 18.yy+xy= 0 7.4y+y= 0 ′′ ′′′ ′ 8.y+y= 019.2y+y+ sin(x)y= 0 ′′ ′′′ ′ 9.y+y+y= 020.4yy+ cos(x)y= 0 ′′ ′ 2′′ ′ 10.yy+y= 021.x y+xyy= 0 ′′ ′ 2′′ ′ 11.y+ 8y+ 2y= 0 22.x y2xy+y= 0