Leveled Literature Titles: List A
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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.



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Topics in Linear Equations
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
n(n1) y+an1y+∙ ∙ ∙+a0y= 0.
The variation of parameters formula and the method of undetermined coefficients are discussed for the associated forced equation
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
n n1 r+an1r+∙ ∙ ∙+a0= 0.
The general solutionyof (1) is constructed as follows.
6.1 Higher Order Linear Equations
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
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.
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
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
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.
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
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).
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