#set document(title: "7.4 Exercises", author: "Jiří Lebl") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 7.4#h(0.6em)Exercises === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/7%3A_Power_series_methods/7.1%3A_Power_Series")[7.1: Power Series] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the power series #math.equation(block: false, alt: "∑ k equals 0 infinity e to the power k x to the power k")[$∑_(k = 0)^(∞) e^(k) x^(k)$] convergent? If so, what is the radius of convergence? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the power series #math.equation(block: false, alt: "∑ k equals 0 infinity k x to the power k")[$∑_(k = 0)^(∞) k x^(k)$] convergent? If so, what is the radius of convergence? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the power series #math.equation(block: false, alt: "∑ k equals 0 infinity k ! x to the power k")[$∑_(k = 0)^(∞) k ! x^(k)$] convergent? If so, what is the radius of convergence? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the power series #math.equation(block: false, alt: "∑ k equals 0 infinity the fraction 1 over open parenthesis 2 k close parenthesis ! open parenthesis x minus 10 close parenthesis to the power k")[$∑_(k = 0)^(∞) frac(1, ( 2 k ) !) attach(( x − 10 ), t: k)$] convergent? If so, what is the radius of convergence? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine the Taylor series for #math.equation(block: false, alt: "sin x")[$sin x$] around the point #math.equation(block: false, alt: "x sub 0 equals π")[$x_(0) = π$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine the Taylor series for #math.equation(block: false, alt: "ln x")[$ln x$] around the point #math.equation(block: false, alt: "x sub 0 equals 1")[$x_(0) = 1$], and find the radius of convergence. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine the Taylor series and its radius of convergence of #math.equation(block: false, alt: "the fraction 1 over 1 plus x")[$frac(1, 1 + x)$] around #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Determine the Taylor series and its radius of convergence of #math.equation(block: false, alt: "the fraction x over 4 minus x squared")[$frac(x, 4 − x^(2))$] around #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. Hint: You will not be able to use the ratio test. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Expand #math.equation(block: false, alt: "x to the power 5 plus 5 x plus 1")[$x^(5) + 5 x + 1$] as a power series around #math.equation(block: false, alt: "x sub 0 equals 5")[$x_(0) = 5$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that the ratio test applies to a series #math.equation(block: false, alt: "∑ k equals 0 infinity a sub k x to the power k")[$∑_(k = 0)^(∞) a_(k) x^(k)$]. Show, using the ratio test, that the radius of convergence of the differentiated series is the same as that of the original series. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Suppose that #math.equation(block: false, alt: "f")[$f$] is an analytic function such that #math.equation(block: false, alt: "f to the power open parenthesis n close parenthesis open parenthesis 0 close parenthesis equals n")[$f^(( n )) ( 0 ) = n$]. Find #math.equation(block: false, alt: "f open parenthesis 1 close parenthesis")[$f ( 1 )$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Is the power series #math.equation(block: false, alt: "∑ n equals 1 infinity open parenthesis 0.1 close parenthesis to the power n x to the power n")[$∑_(n = 1)^(∞) attach(( 0.1 ), t: n) x^(n)$] convergent? If so, what is the radius of convergence? Answer Yes. Radius of convergence is #math.equation(block: false, alt: "10")[$10$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Is the power series #math.equation(block: false, alt: "∑ n equals 1 infinity the fraction n ! over n to the power n x to the power n")[$∑_(n = 1)^(∞) frac(n !, n^(n)) x^(n)$] convergent? If so, what is the radius of convergence? Answer Yes. Radius of convergence is #math.equation(block: false, alt: "e")[$e$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Using the geometric series, expand #math.equation(block: false, alt: "the fraction 1 over 1 minus x")[$frac(1, 1 − x)$] around #math.equation(block: false, alt: "x sub 0 equals 2")[$x_(0) = 2$]. For what #math.equation(block: false, alt: "x")[$x$] does the series converge? Answer #math.equation(block: false, alt: "the fraction 1 over 1 minus x equals minus the fraction 1 over 1 minus open parenthesis 2 minus x close parenthesis")[$frac(1, 1 − x) = − frac(1, 1 − ( 2 − x ))$] so #math.equation(block: false, alt: "the fraction 1 over 1 minus x equals ∑ n equals 0 infinity open parenthesis minus 1 close parenthesis to the power n plus 1 open parenthesis x minus 2 close parenthesis to the power n")[$frac(1, 1 − x) = limits(∑)^(∞)_(n = 0) ( − 1 )^(n + 1) ( x − 2 )^(n)$], which converges for #math.equation(block: false, alt: "1 less than x less than 3")[$1 < x < 3$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Find the Taylor series for #math.equation(block: false, alt: "x to the power 7 e to the power x")[$x^(7) e^(x)$] around #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. Answer #math.equation(block: true, alt: "∑ n equals 7 infinity the fraction 1 over open parenthesis n minus 7 close parenthesis ! x to the power n")[$limits(∑)^(∞)_(n = 7) frac(1, ( n − 7 ) !) x^(n)$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] Imagine #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] are analytic functions such that #math.equation(block: false, alt: "f to the power open parenthesis k close parenthesis open parenthesis 0 close parenthesis equals g to the power open parenthesis k close parenthesis open parenthesis 0 close parenthesis")[$f^(( k )) ( 0 ) = g^(( k )) ( 0 )$] for all large enough #math.equation(block: false, alt: "k")[$k$]. What can you say about #math.equation(block: false, alt: "f open parenthesis x close parenthesis minus g open parenthesis x close parenthesis")[$f ( x ) − g ( x )$]? Answer #math.equation(block: false, alt: "f open parenthesis x close parenthesis minus g open parenthesis x close parenthesis")[$f ( x ) − g ( x )$] is a polynomial. Hint: Use Taylor series. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/7%3A_Power_series_methods/7.2%3A_Series_Solutions_of_Linear_Second_Order_ODEs")[7.2: Series solutions of linear second order ODEs] In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the #math.equation(block: false, alt: "k to the power th")[$k^("th")$] coefficient. #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use power series methods to solve #math.equation(block: false, alt: "y double prime plus y equals 0")[$y^(″) + y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 1")[$x_(0) = 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use power series methods to solve #math.equation(block: false, alt: "y double prime plus 4 x y equals 0")[$y^(″) + 4 x y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use power series methods to solve #math.equation(block: false, alt: "y double prime minus x y equals 0")[$y^(″) − x y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 1")[$x_(0) = 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use power series methods to solve #math.equation(block: false, alt: "y double prime plus x squared y equals 0")[$y^(″) + x^(2) y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The methods work for other orders than second order. Try the methods of this section to solve the first order system #math.equation(block: false, alt: "y prime minus x y equals 0")[$y^(′) − x y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Chebyshev’s equation of order #math.equation(block: false, alt: "p")[$p$]: + Solve #math.equation(block: false, alt: "open parenthesis 1 minus x squared close parenthesis y double prime minus x y prime plus p squared y equals 0")[$( 1 − x^(2) ) y^(″) − x y^(′) + p^(2) y = 0$] using power series methods at #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. + For what #math.equation(block: false, alt: "p")[$p$] is there a polynomial solution? ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a polynomial solution to #math.equation(block: false, alt: "open parenthesis x squared plus 1 close parenthesis y double prime minus 2 x y prime plus 2 y equals 0")[$( x^(2) + 1 ) y^(″) − 2 x y^(′) + 2 y = 0$] using power series methods. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Use power series methods to solve #math.equation(block: false, alt: "open parenthesis 1 minus x close parenthesis y double prime plus y equals 0")[$( 1 − x ) y^(″) + y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. + Use the solution to part a) to find a solution for #math.equation(block: false, alt: "x y double prime plus y equals 0")[$x y^(″) + y = 0$] around the point #math.equation(block: false, alt: "x sub 0 equals 1")[$x_(0) = 1$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Use power series methods to solve #math.equation(block: false, alt: "y double prime plus 2 x cubed y equals 0")[$y^(″) + 2 x^(3) y = 0$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. Answer #math.equation(block: false, alt: "a sub 2 equals 0")[$a_(2) = 0$], #math.equation(block: false, alt: "a sub 3 equals 0")[$a_(3) = 0$], #math.equation(block: false, alt: "a sub 4 equals 0")[$a_(4) = 0$], recurrence relation (for #math.equation(block: false, alt: "k greater than or equal to 5")[$k ≥ 5$]): #math.equation(block: false, alt: "a sub k equals the fraction minus 2 a sub k minus 5 over k open parenthesis k minus 1 close parenthesis")[$a_(k) = frac(− 2 a_(k − 5), k ( k − 1 ))$], so #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals a sub 0 plus a sub 1 x minus the fraction a sub 0 over 10 x to the power 5 minus the fraction a sub 1 over 15 x to the power 6 plus the fraction a sub 0 over 450 x to the power 10 plus the fraction a sub 1 over 825 x to the power 11 minus the fraction a sub 0 over 47250 x to the power 15 minus the fraction a sub 1 over 99000 x to the power 16 plus ⋯")[$y ( x ) = a_(0) + a_(1) x − frac(a_(0), 10) x^(5) − frac(a_(1), 15) x^(6) + frac(a_(0), 450) x^(10) + frac(a_(1), 825) x^(11) − frac(a_(0), 47250) x^(15) − frac(a_(1), 99000) x^(16) + ⋯$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(challenging)] We can also use power series methods in nonhomogeneous equations. + Use power series methods to solve #math.equation(block: false, alt: "y double prime minus x y equals the fraction 1 over 1 minus x")[$y^(″) − x y = frac(1, 1 − x)$] at the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$]. Hint: Recall the geometric series. + Now solve for the initial condition #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$], #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals 0")[$y^(′) ( 0 ) = 0$]. Answer + #math.equation(block: false, alt: "a sub 2 equals the fraction 1 over 2")[$a_(2) = frac(1, 2)$], and for #math.equation(block: false, alt: "k greater than or equal to 1")[$k ≥ 1$] we have #math.equation(block: false, alt: "a sub k equals the fraction a sub k minus 3 plus 1 over k open parenthesis k minus 1 close parenthesis")[$a_(k) = frac(a_(k − 3) + 1, k ( k − 1 ))$], so #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals a sub 0 plus a sub 1 x plus the fraction 1 over 2 x squared plus the fraction a sub 0 plus 1 over 6 x cubed plus the fraction a sub 1 plus 1 over 12 x to the power 4 plus the fraction 3 over 40 x to the power 5 plus the fraction a sub 0 plus 2 over 30 x to the power 6 plus the fraction a sub 1 plus 2 over 42 x to the power 7 plus the fraction 5 over 112 x to the power 8 plus the fraction a sub 0 plus 3 over 72 x to the power 9 plus the fraction a sub 1 plus 3 over 90 x to the power 10 plus ⋯")[$y ( x ) = a_(0) + a_(1) x + frac(1, 2) x^(2) + frac(a_(0) + 1, 6) x^(3) + frac(a_(1) + 1, 12) x^(4) + frac(3, 40) x^(5) + frac(a_(0) + 2, 30) x^(6) + frac(a_(1) + 2, 42) x^(7) + frac(5, 112) x^(8) + frac(a_(0) + 3, 72) x^(9) + frac(a_(1) + 3, 90) x^(10) + ⋯$] + #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals the fraction 1 over 2 x squared plus the fraction 1 over 6 x cubed plus the fraction 1 over 12 x to the power 4 plus the fraction 3 over 40 x to the power 5 plus the fraction 1 over 15 x to the power 6 plus the fraction 1 over 21 x to the power 7 plus the fraction 5 over 112 x to the power 8 plus the fraction 1 over 24 x to the power 9 plus the fraction 1 over 30 x to the power 10 plus ⋯")[$y ( x ) = frac(1, 2) x^(2) + frac(1, 6) x^(3) + frac(1, 12) x^(4) + frac(3, 40) x^(5) + frac(1, 15) x^(6) + frac(1, 21) x^(7) + frac(5, 112) x^(8) + frac(1, 24) x^(9) + frac(1, 30) x^(10) + ⋯$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Attempt to solve #math.equation(block: false, alt: "x squared y double prime minus y equals 0")[$x^(2) y^(″) − y = 0$] at #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$] using the power series method of this section (#math.equation(block: false, alt: "x sub 0")[$x_(0)$] is a singular point). Can you find at least one solution? Can you find more than one solution? Answer Applying the method of this section directly we obtain #math.equation(block: false, alt: "a sub k equals 0")[$a_(k) = 0$] for all #math.equation(block: false, alt: "k")[$k$] and so #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals 0")[$y ( x ) = 0$] is the only solution we find. ] === #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/7%3A_Power_series_methods/7.3%3A_Singular_Points_and_the_Method_of_Frobenius")[7.3: Singular points and the method of Frobenius] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular (Frobenius-type) solution of #math.equation(block: false, alt: "x squared y double prime plus x y prime plus open parenthesis 1 plus x close parenthesis y equals 0")[$x^(2) y^(″) + x y^(′) + ( 1 + x ) y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular (Frobenius-type) solution of #math.equation(block: false, alt: "x y double prime minus y equals 0")[$x y^(″) − y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular (Frobenius-type) solution of #math.equation(block: false, alt: "y double prime plus the fraction 1 over x y prime minus x y equals 0")[$y^(″) + frac(1, x) y^(′) − x y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "2 x y double prime plus y prime minus x squared y equals 0")[$2 x y^(″) + y^(′) − x^(2) y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x squared y double prime minus x y prime minus y equals 0")[$x^(2) y^(″) − x y^(′) − y = 0$]. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the following equations classify the point #math.equation(block: false, alt: "x equals 0")[$x = 0$] as ordinary, regular singular, or singular but not regular singular. + #math.equation(block: false, alt: "x squared open parenthesis 1 plus x squared close parenthesis y double prime plus x y equals 0")[$x^(2) ( 1 + x^(2) ) y^(″) + x y = 0$] + #math.equation(block: false, alt: "x squared y double prime plus y prime plus y equals 0")[$x^(2) y^(″) + y^(′) + y = 0$] + #math.equation(block: false, alt: "x y double prime plus x cubed y prime plus y equals 0")[$x y^(″) + x^(3) y^(′) + y = 0$] + #math.equation(block: false, alt: "x y double prime plus x y prime minus e to the power x y equals 0")[$x y^(″) + x y^(′) − e^(x) y = 0$] + #math.equation(block: false, alt: "x squared y double prime plus x squared y prime plus x squared y equals 0")[$x^(2) y^(″) + x^(2) y^(′) + x^(2) y = 0$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the following equations classify the point #math.equation(block: false, alt: "x equals 0")[$x = 0$] as ordinary, regular singular, or singular but not regular singular. + #math.equation(block: false, alt: "y double prime plus y equals 0")[$y^(″) + y = 0$] + #math.equation(block: false, alt: "x cubed y double prime plus open parenthesis 1 plus x close parenthesis y equals 0")[$x^(3) y^(″) + ( 1 + x ) y = 0$] + #math.equation(block: false, alt: "x y double prime plus x to the power 5 y prime plus y equals 0")[$x y^(″) + x^(5) y^(′) + y = 0$] + #math.equation(block: false, alt: "sin open parenthesis x close parenthesis y double prime minus y equals 0")[$sin ( x ) y^(″) − y = 0$] + #math.equation(block: false, alt: "cos open parenthesis x close parenthesis y double prime minus sin open parenthesis x close parenthesis y equals 0")[$cos ( x ) y^(″) − sin ( x ) y = 0$] Answer + ordinary, + singular but not regular singular, + regular singular, + regular singular, + ordinary. ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the general solution of #math.equation(block: false, alt: "x squared y double prime minus y equals 0")[$x^(2) y^(″) − y = 0$]. Answer #math.equation(block: true, alt: "y equals A x to the power the fraction 1 plus the square root of 5 over 2 plus B x to the power the fraction 1 minus the square root of 5 over 2")[$y = A x^(frac(1 + sqrt(5), 2)) + B x^(frac(1 − sqrt(5), 2))$] ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find a particular solution of #math.equation(block: false, alt: "x squared y double prime plus open parenthesis x minus the fraction 3 over 4 close parenthesis y equals 0")[$x^(2) y^(″) + ( x − frac(3, 4) ) y = 0$]. Answer #math.equation(block: false, alt: "y equals x to the power 3 / 2 ∑ k equals 0 infinity the fraction open parenthesis minus 1 close parenthesis to the power minus 1 over k ! open parenthesis k plus 2 close parenthesis ! x to the power k")[$y = x^(3 / 2) limits(∑)^(∞)_(k = 0) frac(( − 1 )^(− 1), k ! ( k + 2 ) !) x^(k)$] (Note that for convenience we did not pick #math.equation(block: false, alt: "a sub 0 equals 1")[$a_(0) = 1$].) ] #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[(tricky)] Find the general solution of #math.equation(block: false, alt: "x squared y double prime minus x y prime plus y equals 0")[$x^(2) y^(″) − x y^(′) + y = 0$]. Answer #math.equation(block: true, alt: "y equals A x plus B x ln open parenthesis x close parenthesis")[$y = A x + B x ln ( x )$] ]