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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) 2−y1(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 let2−y1 2(x)the nonhomogeneous. 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)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 ′ ′1x−x′ and (coshxsinh) =x. Forinstance, (coshx) standsfor (e+ewhich) , 2 1x−x evaluates to(e−e), or sinhx. 2 Wronskian detail. Lety1= sinhx,y2= coshx. Then ′ ′ (x) (x)y(x) W=y1(x)y2−y1 2Definition of WronskianW. ′ ′ nh(x)−cosh(x) cosh(x)Substitute fory,y, = sinh(x) si1 1y2,y2. 1x−x2 1x−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 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 growthdecay 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 intevariation 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)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=yh+yp=c1e+c2e+ x xe /follow.2. Details 2′′ Homogeneous solution. Thecharacteristic equationr−1 = 0 fory−y= 0 x−x 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. Substitutiony−y=egives 2d1e+ x xx x d1xe−d1xe=e. Canceleand equate coefficients of powers ofxto find x d1= 1/2. Thenyp=xe /2. x Variation of Parameters Summary. Thehomogeneous solutionyh=c1e+ −x x−x 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 x−x x yp(x) =e edx+dx.e e −2−2 Integration, followed by setting all constants of integration to zero, givesyp(x) = x x xe /2−e /4. x Differences. Thetwo methods give respectivelyyp=xe /2 andyp(x) = x xx xx xe /2−e /4. Thesolutionsyp=xe /2 andyp(x) =xe /2−e /4 differ by the x homogeneous solution−xe /the general solution is4. In both cases, 1 x−x 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 growthdecay 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 y−yduct rule;y ycancels. 1 21 21y2−y yPro1 2 1 2 ′ ′ −cy1)y2/aBothy1,y2satisfy (2). =y1(−by−cy2)/a−(−by1 2 ′ ′ y−y 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 relabelling 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. Ify1≡0, 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=y2−cy1linearity,. ByyFurther,is a solution of (2).y(x0) =y(x0) = 0. By uniqueness of initial value problems,y≡0, 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 (17361813), 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.
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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 twotermed 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.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 ′′Wronskianthe 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= 019.2y+y+ sin(x)y= 0 ′′ ′′′ ′ 9.y+y+y= 020.4y−y+ cos(x)y= 0 ′′ ′ 2′′ ′ 10.y−y+y= 021.x y+xy−y= 0 ′′ ′ 2′′ ′ 11.y+ 8y+ 2y= 0 22.x y−2xy+y= 0
