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# Leveled Literature Titles: List A

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• leçon - matière potentielle : plays
• leçon - matière potentielle : authors
• leçon - matière potentielle : elizabeth goudge
• leçon - matière potentielle : 's drummer
• leçon - matière potentielle : day landings
• leçon - matière potentielle : translator
• leçon - matière potentielle : ace amelia earhart
• leçon - matière potentielle : ancient times
• leçon - matière potentielle : level
• leçon - matière : literature - matière potentielle : literature
• leçon - matière potentielle : maeve binchy
• leçon - matière potentielle : continent
Prentice Hall Leveled Library: Novel Lists A and B Revised Novel Lists A & B as of January 2010
• rudyard kipling 820 sc 0142410314 charlie
• washington irving 970 sc 0374443300 leon
• prose sc 0131166204 othello
• sc 0835935922 david copperfield
• mary shelley 940 sc 0756633303 frankenstein
• prose hc
• shirts 600 sc 0451527747 alice
• prose sc
• sir a. c. doyle
• t.h.
• t. h.
• t.-h.

Subjects

##### Maeve Binchy

Informations

Exrait

Chapter
6
Topics in Linear Equations
Diﬀerential
Developed here is the theory for higher order linear constantcoeﬃcient diﬀerential equations. Besides a basic recipe for the solution of such equations, extensions are developed for the topics of variation of param eters and undetermined coeﬃcients. Enrichment topics include the CauchyEuler diﬀerential equation, the Cauchy kernel for second order linear diﬀerential equations, and a library of special methods for undetermined coeﬃcients methods, the latter hav ing prerequisites of only basic calculus and college algebra. Developed with the library methods is a veriﬁcation of the method of undetermined coecients,viaK¨ummersmethod.
6.1 Higher Order Linear Equations
Developed here is the recipe for higher order linear diﬀerential equations with constant coeﬃcients
(1)
n(n1) y+an1y+∙ ∙ ∙+a0y= 0.
The variation of parameters formula and the method of undetermined coeﬃcients are discussed for the associated forced equation
(2)
n(n1) y+an1y+∙ ∙ ∙+a0y=r(x).
A Recipe for Higher Order Equations
Consider equation (1) withrealcoeﬃcients. Thecharacteristic equa tionof (1) is the polynomial equation
(3)
n n1 r+an1r+∙ ∙ ∙+a0= 0.
The general solutionyof (1) is constructed as follows.
6.1 Higher Order Linear Equations
239
Higher Order Recipe Stage 1. Repeat (I) below for all distinct real rootsr=aof the characteristic k equation (3). Symbolkis the maximum power such that (ra) divides the characteristic polynomial, which means thatkequals the algebraic multiplicity of the rootr=a.
(I)The equationra= 0 is the characteristic equation ofuau= 0, having general solution ax u=u0e . Replaceu0by a polynomial inxwithkarbitrary coeﬃcients. Add the modiﬁed expressionuto the general solutiony.
Higher Order Recipe Stage 2. Repeat (II) below for all distinct complex rootsz=a+ib,b >0, of the characteristic equation (3). Symbolkis the maximum power such k that (rz) divides the characteristic polynomial, which means thatk equals the algebraic multiplicity of the rootr=z.
(II)The equation (rz)(rz) = 0 is the characteristic equation of a second order diﬀerential equation whoseCase 3recipe solution is
ax ax u=u1ecosbx+u2esinbx.
Replace the constantsu1,u2by polynomials inxwithkarbitrary coeﬃcients, a total of 2kcoeﬃcients. Add the modiﬁed expression uto the general solutiony.
Exponential Solutions.Characteristic equation (3) is formally ob (k)k tained from the diﬀerential equation by replacingybyr. This device for remembering how to form the characteristic equation is attributed to Euler, because of the following fact.
Theorem 1 (Euler’s Exponential Substitution) wx Letwbe a real or complex number. The functiony(x) =eis a solution of (1) if and only ifr=wis a root of the characteristic equation (3).
Factorization.According to the fundamental theorem of algebra, equation (3) has exactlynroots, counted according to multiplicity. Some number of the roots are real and the remaining roots appear in complex conjugate pairs. This implies that every characteristic equation has a factored form
k1kqm1mp (ra1)∙ ∙ ∙(raq)Q1(r)∙ ∙ ∙Qp(r) = 0
240
Topics in Linear Diﬀerential Equations
wherea1, . . . ,aqare thedistinct real rootsof the characteristic equa tion of algebraic multiplicitiesk1. . ,, . kq, respectively, andQ1(r), . . . , Qp(r) are the distinct real quadratic factors of the form (rz)(rz), wherezexhausts thedistinct complex rootsz=a+ibwithb >0, having corresponding multiplicitiesm1, . . . ,mp. Some Recipe Details. RecipeStage 1loops on the distinct linear factors while recipeStage 2loops on the distinct real quadratic factors. Theydiﬀerential equation can be expressed inDoperator notation as   k1kqm1mp (Da1)∙ ∙ ∙(Daq)Q1(D)∙ ∙ ∙Qp(D)y= 0.
The recipe is based upon the fact that the general solutionyis the sum of general solution expressions obtained from each distinct factor in this operator form. Speciﬁcally, the general solution of
k+1 (Da)y= 0
k ax is a polynomialu=c0+c1x+∙ ∙ ∙+ckxwithk+ 1 terms timese. ax This fact is proved by the change of variabley=e u, which ﬁnds an k+1 equivalent equationD u= 0, solvable by quadrature.
An Illustration of the Higher Order Recipe. Consider the problem of solving a constant coeﬃcient linear diﬀerential equation (1) of order 11 having factored characteristic equation
3 2 2 2 2 (r2) (r(+ 1) r+ 4) (r+ 4r+ 5) = 0.
To be applied is the recipe for higher order equations. ThenStage 1 loops on the two linear factorsr2 andrwhile+ 1, Stage 2loops on 2 2 the two real quadratic factorsr+ 4 andr+ 4r+ 5. Hand solutions can be organized by a tabular method for generating the general solutiony.
Factor Multiplicity Base Root Base Solution
3 (r2) 3 r= 2 2x u0e
2 (r+ 1) 2 r=1 ∗ −x u e 0
2 2 (r+ 4) 2 r= 0 + 2i u1cos 2x +u2sin 2x
2 (r+ 4r+ 5) 1 r=2 +i ∗ −2x u ecosx 1 ∗ −2x +u esinx 2
Symbolsc1, . . . ,c11will represent arbitrary constants in the general so ∗ ∗ lutiony. Symbolsu0,u,u1,u2,u,uinitially represent constants, but 0 1 2 they will be assigned polynomial expressions, according to root multi
6.1 Higher Order Linear Equations
plicity, as follows.
Root Multiplicity Polynomial Assigned 2 r= 2 3u0=c1+c2x+c3x r=1 2u=c4+c5x 0 r= 0 + 2i2u1=c6+c7x u2=c8+c9x r=2 +i1u=c10 1 u 2=c11 The recipeStage 1andStage 2solutions are added toy, giving 2x∗ −x y=u0e+uin 2x 0e+u1cos 2x+u2s ∗ −2x∗ −2x +u ecosx+u esinx 1 2 2 2x = (c1+c2x+c3x)e x +(c4+c5x)e +(c6+c7x) cos 2x+ (c8+c9x) sin 2x 2x2x +c10ecosx+c11esinx.
241
Computer Algebra System Solution.The systemmaplecan symbolically solve a higher order equation. Below,@is the function composition operator,@@is the repeated composition operator andDis 3 the diﬀerentiation operator. The coding writes the factors of (r2) (r+ 2 2 2 2 3 2 1) (D(+ 4) D+ 4D+ 5) as diﬀerential operators (D2) , (D,+ 1) 2 2 2 (D+ 4) ,D+ 4DThen the diﬀerential equation is the composition+ 5. of the component factors.
id:=x>x; F1:=(D2*id) @@ 3; F2:=(D+id) @@ 2; F3:=(D@D+4*id) @@ 2; F4:=D@D+4*D+5*id; de:=(F1@F2@F3@F4)(y)(x)=0: dsolve({de},y(x));
Variation of Parameters Formula
ThePicardLindel¨oftheoremimpliesauniquesolutiondenedon(−∞,) for the initial value problem
(4)
n(n1) y+an1y+∙ ∙ ∙+a0y= 0, (n2) (n1) y(0) =∙ ∙ ∙=y(0) = 0, y(0) = 1.
The unique solution is calledCauchy’s kernel, writtenK(x). ′′′ ′′ To illustrate, Cauchy’s kernelK(x) foryy= 0 is obtained from x its general solutiony=c1+c2x+c3eby computing the values of the
242
Topics in Linear Diﬀerential Equations
′ ′′ constants from initial conditionsy(0) = 0,y(0) = 0,y(0) = 1, giving x K(x) =ex1.
Theorem 2 (Higher Order Variation of Parameters) n(n1) Lety+an1y+∙ ∙ ∙+a0y=r(x)have constant coeﬃcientsa0. . ,, . an1and continuous forcing termr(x). Denote byK(x)Cauchy’s kernel for the homogeneous diﬀerential equation. Then a particular solution is given by thevariation of parameters formula Z x (5)yp(x) =K(xu)r(u)du. 0 (n1) This solution has zero initial conditionsy(0) =∙ ∙ ∙=y(0) = 0. R x Proof:Deﬁney(x) =K(xu)r(u)duby the 2variable chain. Compute R 0 x rule applied toF(x, y) =K(yu)r(u)duthe formulae 0 y(x) =F(x, x) R x =K(xu)r(u)du, 0 y(x) =Fx(x, x,) +Fy(x, x) R x =K(xx)r(x) +K(xu)r(u)du R 0 x = 0 +K(xu)r(u)du. 0 The process can be continued to obtain for 0p < n1 the general relation Z x (p) (p) y(x) =Kr(u)du. 0 (n1) The relation justiﬁes the initial conditionsy(0) =∙ ∙ ∙=y(0) = 0, because each integral is zero atx= 0. Takep=n1 and diﬀerentiate once again to give Z x (n) (n1) (n) y(x) =K(xx)r(x) +Kr(u)du. 0 (n1) BecauseK(0) = 1, this relation implies  ! n1Zn1 x X X (n) (p) (n) (p) y+apy=r(x) +K(xu) +apK(xu)r(u)du. 0 p=0p=0
The sum under the integrand on the right is zero, because Cauchy’s kernel satisﬁes the homogeneous diﬀerential equation. This provesy(x) satisﬁes the nonhomogeneous diﬀerential equation. The proof is complete.
Undetermined
Coeﬃcients Method
The method applies to higher order nonhomogeneous diﬀerential equa tions (n1) (6)y+an1y+∙ ∙ ∙+a0y=r(x).
It ﬁnds a particular solutionypof (6)withoutthe integration steps present in variation of parameters. The requirements and limitations:
6.1 Higher Order Linear Equations
1coeﬃcients on the left side of (6) are constant.. The 2. The functionr(x)is a sum of constants times atoms.
Anatomis a term having one of the forms
m m ax m m m ax x , x e , xcosbx, xsinebx, x cosbx
or
243
m ax x esinbx.
The symbolsaandbare real constants, withb >0. Symbolm0 is an integer. AtomsAandBare calledrelated atomsif their successive derivative formulae contain a common atom.
Higher Order Basic Trial Solution Method 1diﬀerentiate the atoms of. Repeatedly r(x)until no new atoms ap pear. Multiply the distinct atoms so found byundetermined co eﬃcientsd1,d2, . . . ,dk, then add to deﬁne atrial solutiony. (n) (n1) 2.Fixup rulethe homogeneous equation: if y+an1y+∙ ∙ ∙+ a0y= 0has solutionyhcontaining an atomAwhich appears in the trial solutiony, then replace eachrelated atomBinybyxB (other atoms appearing inyare unchanged). Repeat the ﬁxup rule untilycontains no atom ofyhmodiﬁed expression. The yis called thecorrected trial solution. (n) (n1) 3. Substituteyinto the diﬀerential equationy+an1y+∙ ∙ ∙+ a0y=r(x). Match atoms left and right to write out linear algebraic equations for the undetermined coeﬃcientsd1,d2, . . . ,dk. 4the equations. The trial solution. Solve ywith evaluated coeﬃcients d1,d2. . ,, . dkbecomes the particular solutionyp.
Higher Order Undetermined Coeﬃcients Illustration. We will solve ′′′ ′′x yy=xe+ 2x+ 1 + 3 sinx, verifying 3 1 1 3 3 2 3x2x yp(x) =xx2xe+x e+ cosx+ sinx. 2 3 2 2 2
Solution: x Test Applicabilityright side. The r(x) =xe+ 2x+ 1 + 3 sinxis a sum of x terms constructed from the atomsxe,x, 1, sinx. The left side has constant coeﬃcients. Therefore, the method of undetermined coeﬃcients applies to ﬁnd a particular solutionyp. Trial Solution. The atoms ofr(xThe dis) are subjected to diﬀerentiation. x x tinct atoms so found are 1,x,e,xe, cosx, sinx(drop coeﬃcients to identify new atoms). The initial trial solution is the expression x x y=d1(1) +d2(x) +d3(e) +d4(xe) +d5(cosx) +d6(sinx).
244
Topics in Linear Diﬀerential Equations
x′′′ ′′x The general solutionyh=c1+c2x+c3eofyy= 0 has atoms 1,x,e, all 2 of which appear in the trial solutionyatoms 1,. Multiply related xinybyx to eliminate duplicate atoms 1,xwhich appear inyhmultiply related. Then x x x atomse,xeinybyxto eliminate the duplicate atomewhich appears in yhother atoms cos. The x, sinxinyare unaﬀected by the ﬁxup rule, because they are unrelated to atoms ofyhﬁnal trial solution is. The
2 3x2x y=d1(x) +d2(x) +d3(xe) +d4(x e) +d5(cosx) +d6(sinx).
′′′ ′′ Equations. To substitute the trial solutionyintoyyrequires formulae ′ ′′ ′′′ fory,y,y:
2xx x 2x y= 2d1x+ 3d2x+d3e x+d3e+ 2d4xe+d4x e d5sin(x) +d6cos(x), ′′x xx x 2x y= 2d1+ 6d2x+d3e x+ 2d3e+ 2d4e+ 4d4xe+d4x e d5cos(x)d6sin(x), ′′′x xx x 2x y= 6d2+d3e x+ 3d3e+ 6d4e+ 6d4xe+d4x e +d5sin(x)d6cos(x) Then ′′′ ′′ r(x) =yyThe given equation. x x = 6d22d16d2x+ (d3+ 4d4)e+ 2d4xeSubstitute, then + (d5d6) cos(x) + (d5+d6) sin(x)collect like terms.
x Also,r(x)1 + 2x+xe+ 3 sinxof atoms on the left and right. Coeﬃcients must match. Writing out the matches gives the equations
2d1+ 6d2= 1, 6d2= 2, d3+ 4d4= 0, 2d4= 1, d5d6= 0, d5+d6= 3.
Solveﬁrst four equations can be solved by backsubstitution to give. The d2=1/3,d1=3/2,d4= 1/2,d3=2. The last two equations are solved by elimination or Cramer’s rule to gived5= 3/2,d6= 3/2. Reportyp. The trial solutionywith evaluated coeﬃcientsd1. . ,, . d6becomes
3 1 1 3 3 2 3x2x yp(x) =xx2xe+x e+ cosx+ sinx. 2 3 2 2 2
Exercises 6.1
Higher Order Recipe Factored. Solve the higher order equation with the given characteristic equation. Use the higher order recipe and display a table of distinct roots, multiplicities and base solutions. Verify the gen
eral solutionywith a computer algebra system, if possible.
2 1.(r1)(r+ 2)(r3) = 0
2 2.(r1) (r+ 2)(r+ 3) = 0