#set document(title: "7.2 Series Solutions of Linear Second Order ODEs", 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"))


    
== 7.2#h(0.6em)Series Solutions of Linear Second Order ODEs


    Suppose we have a linear second order homogeneous ODE of the form


    #math.equation(block: true, alt: "p open parenthesis x close parenthesis y double prime plus q open parenthesis x close parenthesis y prime plus r open parenthesis x close parenthesis y equals 0")[$p ( x ) y^(″) + q ( x ) y^(′) + r ( x ) y = 0$]


    Suppose that #math.equation(block: false, alt: "p open parenthesis x close parenthesis")[$p ( x )$], #math.equation(block: false, alt: "q open parenthesis x close parenthesis")[$q ( x )$], and #math.equation(block: false, alt: "r open parenthesis x close parenthesis")[$r ( x )$] are polynomials. We will try a solution of the form


    #math.equation(block: true, alt: "y equals ∑ k equals 0 infinity a sub k open parenthesis x minus x sub o close parenthesis squared")[$y = ∑_(k = 0)^(∞) a_(k) attach(( x − x_(o) ), t: 2)$]


    and solve for the #math.equation(block: false, alt: "a sub k")[$a_(k)$] to try to obtain a solution defined in some interval around #math.equation(block: false, alt: "x sub o")[$x_(o)$].


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Ordinary and Singular Points]


      The point #math.equation(block: false, alt: "x sub o")[$x_(o)$] is called an #emph[ordinary point] if #math.equation(block: false, alt: "p open parenthesis x sub o close parenthesis not equal to 0")[$p ( x_(o) ) ≠ 0$] in linear second order homogeneous ODE of the form in Equation \\ref{eq1}. That is, the functions


      #math.equation(block: true, alt: "the fraction q open parenthesis x close parenthesis over p open parenthesis x close parenthesis and the fraction r open parenthesis x close parenthesis over p open parenthesis x close parenthesis")[$frac(q ( x ), p ( x )) #h(1em) "and" #h(1em) frac(r ( x ), p ( x ))$]


      are defined for #math.equation(block: false, alt: "x")[$x$] near #math.equation(block: false, alt: "x sub o")[$x_(o)$].


      If #math.equation(block: false, alt: "p open parenthesis x sub 0 close parenthesis equals 0")[$p ( x_(0) ) = 0$], then we say #math.equation(block: false, alt: "x sub o")[$x_(o)$] is a #emph[singular point].
]


    Handling singular points is harder than ordinary points and so we now focus only on ordinary points.


    #examplebox("Example 1")[Expansion around an Ordinary Point][
Let us start with a very simple example


        #math.equation(block: true, alt: "y double prime minus y equals 0")[$y^(″) − y = 0$]


        Let us try a power series solution near #math.equation(block: false, alt: "x sub o equals 0")[$x_(o) = 0$], which is an ordinary point.


      
      #solutionbox[
Every point is an ordinary point in fact, as the equation is constant coefficient. We already know we should obtain exponentials or the hyperbolic sine and cosine, but let us pretend we do not know this.


        We try


        #math.equation(block: true, alt: "y equals ∑ k equals 0 infinity a sub k x to the power k")[$y = ∑_(k = 0)^(∞) a_(k) x^(k)$]


        If we differentiate, the #math.equation(block: false, alt: "k equals 0")[$k = 0$] term is a constant and hence disappears. We therefore get


        #math.equation(block: true, alt: "y prime equals ∑ k equals 1 infinity k a sub k x to the power k minus 1")[$y^(′) = ∑_(k = 1)^(∞) k a_(k) x^(k − 1)$]


        We differentiate yet again to obtain (now the #math.equation(block: false, alt: "k equals 1")[$k = 1$] term disappears)


        #math.equation(block: true, alt: "y double prime equals ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2")[$y^(″) = ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2)$]


        We reindex the series (replace #math.equation(block: false, alt: "k")[$k$] with #math.equation(block: false, alt: "k plus 2")[$k + 2$] ) to obtain


        #math.equation(block: true, alt: "y double prime equals ∑ k equals 0 infinity open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 x to the power k")[$y^(″) = ∑_(k = 0)^(∞) ( k + 2 ) ( k + 1 ) a_(k + 2) x^(k)$]


        Now we plug #math.equation(block: false, alt: "y")[$y$] and #math.equation(block: false, alt: "y double prime")[$y^(″)$] into the differential equation.


        #math.equation(block: true, alt: "0 equals y double prime minus y; equals open parenthesis ∑ k equals 0 infinity open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 x to the power k close parenthesis minus open parenthesis ∑ k equals 0 infinity a sub k x to the power k close parenthesis; equals ∑ k equals 0 infinity open parenthesis open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 x to the power k minus a sub k x to the power k close parenthesis; equals ∑ k equals 0 infinity open parenthesis open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 minus a sub k close parenthesis x to the power k .")[$0 = y^(″) − y \ = ( ∑_(k = 0)^(∞) ( k + 2 ) ( k + 1 ) a_(k + 2) x^(k) ) − ( ∑_(k = 0)^(∞) a_(k) x^(k) ) \ = ∑_(k = 0)^(∞) ( ( k + 2 ) ( k + 1 ) a_(k + 2) x^(k) − a_(k) x^(k) ) \ = ∑_(k = 0)^(∞) ( ( k + 2 ) ( k + 1 ) a_(k + 2) − a_(k) ) x^(k) .$]


        As #math.equation(block: false, alt: "y double prime minus y")[$y^(″) − y$] is supposed to be equal to 0, we know that the coefficients of the resulting series must be equal to 0. Therefore,


        #math.equation(block: true, alt: "open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 minus a sub k equals 0")[$( k + 2 ) ( k + 1 ) a_(k + 2) − a_(k) = 0$]


        or


        #math.equation(block: true, alt: "a sub k plus 2 equals the fraction a sub k over open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis .")[$a_(k + 2) = frac(a_(k), ( k + 2 ) ( k + 1 )) .$]


        The above equation is called a#emph[recurrence relation] for the coefficients of the power series. It did not matter what #math.equation(block: false, alt: "a sub 0")[$a_(0)$] or #math.equation(block: false, alt: "a sub 1")[$a_(1)$] was. They can be arbitrary. But once we pick #math.equation(block: false, alt: "a sub 0")[$a_(0)$] and #math.equation(block: false, alt: "a sub 1")[$a_(1)$], then all other coefficients are determined by the recurrence relation.


        Let us see what the coefficients must be. First, #math.equation(block: false, alt: "a sub 0")[$a_(0)$] and #math.equation(block: false, alt: "a sub 1")[$a_(1)$] are arbitrary


        #math.equation(block: true, alt: "a sub 2 equals the fraction a sub 0 over 2 , a sub 3 equals the fraction a sub 1 over open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis , a sub 4 equals the fraction a sub 2 over open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis equals the fraction a sub 0 over open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis , a sub 5 equals the fraction a sub 3 over open parenthesis 5 close parenthesis open parenthesis 4 close parenthesis equals the fraction a sub 1 over open parenthesis 5 close parenthesis open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis , …")[$a_(2) = frac(a_(0), 2) , #h(1em) a_(3) = frac(a_(1), ( 3 ) ( 2 )) , #h(1em) a_(4) = frac(a_(2), ( 4 ) ( 3 )) = frac(a_(0), ( 4 ) ( 3 ) ( 2 )) , #h(1em) a_(5) = frac(a_(3), ( 5 ) ( 4 )) = frac(a_(1), ( 5 ) ( 4 ) ( 3 ) ( 2 )) , #h(1em) …$]


        So we note that for even #math.equation(block: false, alt: "k")[$k$], that is #math.equation(block: false, alt: "k equals 2 n")[$k = 2 n$] we get


        #math.equation(block: true, alt: "a sub k equals a sub 2 n equals the fraction a sub o over open parenthesis 2 n close parenthesis !")[$a_(k) = a_(2 n) = frac(a_(o), ( 2 n ) !)$]


        and for odd #math.equation(block: false, alt: "k")[$k$] that is #math.equation(block: false, alt: "k equals 2 n plus 1")[$k = 2 n + 1$] we have


        #math.equation(block: true, alt: "a sub k equals a sub 2 n plus 1 equals the fraction a sub 1 over open parenthesis 2 n plus 1 close parenthesis !")[$a_(k) = a_(2 n + 1) = frac(a_(1), ( 2 n + 1 ) !)$]


        Let us write down the series


        #math.equation(block: true, alt: "y equals ∑ k equals 0 infinity a sub k x to the power k equals ∑ n equals 0 infinity open parenthesis the fraction a sub 0 over open parenthesis 2 n close parenthesis ! x to the power 2 n plus the fraction a sub 1 over open parenthesis 2 n plus 1 close parenthesis ! x to the power 2 n plus 1 close parenthesis equals a sub 0 ∑ n equals 0 infinity the fraction 1 over open parenthesis 2 n close parenthesis ! x to the power 2 n plus a sub 1 ∑ n equals 0 infinity the fraction 1 over open parenthesis 2 n plus 1 close parenthesis ! x to the power 2 n plus 1 .")[$y = ∑_(k = 0)^(∞) a_(k) x^(k) = ∑_(n = 0)^(∞) ( frac(a_(0), ( 2 n ) !) x^(2 n) + frac(a_(1), ( 2 n + 1 ) !) x^(2 n + 1) ) = a_(0) ∑_(n = 0)^(∞) frac(1, ( 2 n ) !) x^(2 n) + a_(1) ∑_(n = 0)^(∞) frac(1, ( 2 n + 1 ) !) x^(2 n + 1) .$]


        We recognize the two series as the hyperbolic sine and cosine. Therefore,


        #math.equation(block: true, alt: "y equals a sub o cosh x plus a sub 1 sinh x")[$y = a_(o) "cosh" x + a_(1) "sinh" x$]


        Of course, in general we will not be able to recognize the series that appears, since usually there will not be any elementary function that matches it. In that case we will be content with the series.
]
]


    #examplebox("Example 2")[][
Let us do a more complex example. Suppose we wish to solve #emph[Airy’s equation]#math.equation(block: false, alt: "to the power 1")[$1$], that is


        #math.equation(block: true, alt: "y double prime minus x y equals 0")[$y^(″) − x y = 0$]


        near the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$], which is an ordinary point.


        We try


        #math.equation(block: true, alt: "y equals ∑ k equals 0 infinity a sub k x to the power k")[$y = ∑_(k = 0)^(∞) a_(k) x^(k)$]


        We differentiate twice (as above) to obtain


        #math.equation(block: true, alt: "y double prime equals ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2")[$y^(″) = ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2)$]


        We plug #math.equation(block: false, alt: "y")[$y$] into the equation


        #math.equation(block: true, alt: "0 equals y double prime minus x y equals open parenthesis ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2 close parenthesis minus x open parenthesis ∑ k equals 0 infinity a sub k x to the power k close parenthesis; equals open parenthesis ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2 close parenthesis minus open parenthesis ∑ k equals 0 infinity a sub k x to the power k plus 1 close parenthesis .")[$0 = y^(″) − x y = ( ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2) ) − x ( ∑_(k = 0)^(∞) a_(k) x^(k) ) \ = ( ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2) ) − ( ∑_(k = 0)^(∞) a_(k) x^(k + 1) ) .$]


        We reindex to make things easier to sum


        #math.equation(block: true, alt: "0 equals y double prime minus x y equals open parenthesis 2 a sub 2 plus ∑ k equals 1 infinity open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 x to the power k close parenthesis minus open parenthesis ∑ k equals 1 infinity a sub k minus 1 x to the power k close parenthesis .; equals 2 a sub 2 plus ∑ k equals 1 infinity open parenthesis open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 minus a sub k minus 1 close parenthesis x to the power k .")[$0 = y^(″) − x y = ( 2 a_(2) + ∑_(k = 1)^(∞) ( k + 2 ) ( k + 1 ) a_(k + 2) x^(k) ) − ( ∑_(k = 1)^(∞) a_(k − 1) x^(k) ) . \ = 2 a_(2) + ∑_(k = 1)^(∞) ( ( k + 2 ) ( k + 1 ) a_(k + 2) − a_(k − 1) ) x^(k) .$]


        Again #math.equation(block: false, alt: "y double prime minus x y")[$y^(″) − x y$] is supposed to be 0 so first we notice that #math.equation(block: false, alt: "a sub 2 equals 0")[$a_(2) = 0$] and also


        #math.equation(block: true, alt: "open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 minus a sub k minus 1 equals 0 , or a sub k plus 2 equals the fraction a sub k minus 1 over open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis .")[$( k + 2 ) ( k + 1 ) a_(k + 2) − a_(k − 1) = 0 , #h(2em) "or" #h(2em) a_(k + 2) = frac(a_(k − 1), ( k + 2 ) ( k + 1 )) .$]


        Now we jump in steps of three. First we notice that since #math.equation(block: false, alt: "a sub 2 equals 0")[$a_(2) = 0$] we must have that, #math.equation(block: false, alt: "a sub 5 equals 0")[$a_(5) = 0$], #math.equation(block: false, alt: "a sub 8 equals 0")[$a_(8) = 0$], #math.equation(block: false, alt: "a sub 11 equals 0")[$a_(11) = 0$], etc #math.equation(block: false, alt: "…")[$…$]. In general #math.equation(block: false, alt: "a sub 3 n plus 2 equals 0")[$a_(3 n + 2) = 0$]. The constants #math.equation(block: false, alt: "a sub 0")[$a_(0)$] and #math.equation(block: false, alt: "a sub 1")[$a_(1)$] are arbitrary and we obtain


        #math.equation(block: true, alt: "a sub 3 equals the fraction a sub 0 over open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis , a sub 4 equals the fraction a sub 1 over open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis , a sub 6 equals the fraction a sub 3 over open parenthesis 6 close parenthesis open parenthesis 5 close parenthesis equals the fraction a sub 0 over open parenthesis 6 close parenthesis open parenthesis 5 close parenthesis open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis , a sub 7 equals the fraction a sub 4 over open parenthesis 7 close parenthesis open parenthesis 6 close parenthesis equals the fraction a sub 1 over open parenthesis 7 close parenthesis open parenthesis 6 close parenthesis open parenthesis 4 close parenthesis open pare")[$a_(3) = frac(a_(0), ( 3 ) ( 2 )) , #h(1em) a_(4) = frac(a_(1), ( 4 ) ( 3 )) , #h(1em) a_(6) = frac(a_(3), ( 6 ) ( 5 )) = frac(a_(0), ( 6 ) ( 5 ) ( 3 ) ( 2 )) , #h(1em) a_(7) = frac(a_(4), ( 7 ) ( 6 )) = frac(a_(1), ( 7 ) ( 6 ) ( 4 ) ( 3 )) , #h(1em) …$]


        For #math.equation(block: false, alt: "a sub k")[$a_(k)$] where #math.equation(block: false, alt: "k")[$k$] is a multiple of #math.equation(block: false, alt: "3")[$3$], that is #math.equation(block: false, alt: "k equals 3 n")[$k = 3 n$] we notice that


        #math.equation(block: true, alt: "a sub 3 n equals the fraction a sub 0 over open parenthesis 2 close parenthesis open parenthesis 3 close parenthesis open parenthesis 5 close parenthesis open parenthesis 6 close parenthesis ⋯ open parenthesis 3 n minus 1 close parenthesis open parenthesis 3 n close parenthesis .")[$a_(3 n) = frac(a_(0), ( 2 ) ( 3 ) ( 5 ) ( 6 ) ⋯ ( 3 n − 1 ) ( 3 n )) .$]


        For #math.equation(block: false, alt: "a sub k")[$a_(k)$] where #math.equation(block: false, alt: "k equals 3 n plus 1")[$k = 3 n + 1$], we notice


        #math.equation(block: true, alt: "a sub 3 n plus 1 equals the fraction a sub 1 over open parenthesis 3 close parenthesis open parenthesis 4 close parenthesis open parenthesis 6 close parenthesis open parenthesis 7 close parenthesis ⋯ open parenthesis 3 n close parenthesis open parenthesis 3 n plus 1 close parenthesis .")[$a_(3 n + 1) = frac(a_(1), ( 3 ) ( 4 ) ( 6 ) ( 7 ) ⋯ ( 3 n ) ( 3 n + 1 )) .$]


        In other words, if we write down the series for #math.equation(block: false, alt: "y")[$y$] we notice that it has two parts


        #math.equation(block: true, alt: "y equals open parenthesis a sub 0 plus the fraction a sub 0 over 6 x cubed plus the fraction a sub 0 over 180 x to the power 6 plus ⋯ plus the fraction a sub 0 over open parenthesis 2 close parenthesis open parenthesis 3 close parenthesis open parenthesis 5 close parenthesis open parenthesis 6 close parenthesis ⋯ open parenthesis 3 n minus 1 close parenthesis open parenthesis 3 n close parenthesis x to the power 3 n plus ⋯ close parenthesis; plus open parenthesis a sub 1 x plus the fraction a sub 1 over 12 x to the power 4 plus the fraction a sub 1 over 504 x to the power 7 plus ⋯ plus the fraction a sub 1 over open parenthesis 3 close parenthesis open parenthesis 4 close parenthesis open parenthesis 6 close parenthesis open parenthesis 7 close parenthesis ⋯ open parenthesis 3 n close pare")[$y = ( a_(0) + frac(a_(0), 6) x^(3) + frac(a_(0), 180) x^(6) + ⋯ + frac(a_(0), ( 2 ) ( 3 ) ( 5 ) ( 6 ) ⋯ ( 3 n − 1 ) ( 3 n )) x^(3 n) + ⋯ ) \ + ( a_(1) x + frac(a_(1), 12) x^(4) + frac(a_(1), 504) x^(7) + ⋯ + frac(a_(1), ( 3 ) ( 4 ) ( 6 ) ( 7 ) ⋯ ( 3 n ) ( 3 n + 1 )) x^(3 n + 1) + ⋯ ) \ = a_(0) ( 1 + frac(1, 6) x^(3) + frac(1, 180) x^(6) + ⋯ + frac(1, ( 2 ) ( 3 ) ( 5 ) ( 6 ) ⋯ ( 3 n − 1 ) ( 3 n )) x^(3 n) + ⋯ ) \ + a_(1) ( x + frac(1, 12) x^(4) + frac(1, 504) x^(7) + ⋯ + frac(1, ( 3 ) ( 4 ) ( 6 ) ( 7 ) ⋯ ( 3 n ) ( 3 n + 1 )) x^(3 n + 1) + ⋯ ) .$]


        We define


        #math.equation(block: true, alt: "y sub 1 open parenthesis x close parenthesis equals 1 plus the fraction 1 over 6 x cubed plus the fraction 1 over 180 x to the power 6 plus ⋯ plus the fraction 1 over open parenthesis 2 close parenthesis open parenthesis 3 close parenthesis open parenthesis 5 close parenthesis open parenthesis 6 close parenthesis ⋯ open parenthesis 3 n minus 1 close parenthesis open parenthesis 3 n close parenthesis x to the power 3 n plus ⋯ ,; y sub 2 open parenthesis x close parenthesis equals x plus the fraction 1 over 12 x to the power 4 plus the fraction 1 over 504 x to the power 7 plus ⋯ plus the fraction 1 over open parenthesis 3 close parenthesis open parenthesis 4 close parenthesis open parenthesis 6 close parenthesis open parenthesis 7 close parenthesis ⋯ open parenthesis 3 n close parenthesis op")[$y_(1) ( x ) = 1 + frac(1, 6) x^(3) + frac(1, 180) x^(6) + ⋯ + frac(1, ( 2 ) ( 3 ) ( 5 ) ( 6 ) ⋯ ( 3 n − 1 ) ( 3 n )) x^(3 n) + ⋯ , \ y_(2) ( x ) = x + frac(1, 12) x^(4) + frac(1, 504) x^(7) + ⋯ + frac(1, ( 3 ) ( 4 ) ( 6 ) ( 7 ) ⋯ ( 3 n ) ( 3 n + 1 )) x^(3 n + 1) + ⋯ ,$]


        and write the general solution to the equation as #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals a sub 0 y sub 1 open parenthesis x close parenthesis plus a sub 1 y sub 2 open parenthesis x close parenthesis")[$y ( x ) = a_(0) y_(1) ( x ) + a_(1) y_(2) ( x )$]. Notice from the power series that #math.equation(block: false, alt: "y sub 1 open parenthesis 0 close parenthesis equals 1")[$y_(1) ( 0 ) = 1$] and #math.equation(block: false, alt: "y sub 2 open parenthesis 0 close parenthesis equals 0")[$y_(2) ( 0 ) = 0$]. Also, #math.equation(block: false, alt: "y 1 ′ open parenthesis 0 close parenthesis equals 0")[$y_(1)^(′) ( 0 ) = 0$] and #math.equation(block: false, alt: "y 2 ′ open parenthesis 0 close parenthesis equals 1")[$y_(2)^(′) ( 0 ) = 1$]. Therefore #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] is a solution that satisfies the initial conditions #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals a sub 0")[$y ( 0 ) = a_(0)$] and #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals a sub 1")[$y^(′) ( 0 ) = a_(1)$].


        #figure(figph[Graphs of two solutions to Airy's equations.  They look pretty different], alt: "Graphs of two solutions to Airy's equations. They look pretty different", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: The two solutions #math.equation(block: false, alt: "y sub 1")[$y_(1)$] and #math.equation(block: false, alt: "y sub 2")[$y_(2)$] to Airy's equation.])


        The functions #math.equation(block: false, alt: "y sub 1")[$y_(1)$] and #math.equation(block: false, alt: "y sub 2")[$y_(2)$] cannot be written in terms of the elementary functions that you know. See Figure #math.equation(block: false, alt: "1")[$1$] for the plot of the solutions #math.equation(block: false, alt: "y sub 1")[$y_(1)$] and #math.equation(block: false, alt: "y sub 2")[$y_(2)$]. These functions have many interesting properties. For example, they are oscillatory for negative #math.equation(block: false, alt: "x")[$x$] (like solutions to #math.equation(block: false, alt: "y double prime plus y equals 0")[$y^(″) + y = 0$]) and for positive #math.equation(block: false, alt: "x")[$x$] they grow without bound (like solutions to #math.equation(block: false, alt: "y double prime minus y equals 0")[$y^(″) − y = 0$]).
]


    Sometimes a solution may turn out to be a polynomial.


    
      The even series solution y₁ of Hermite's equation y″ − 2xy′ + 2ny = 0, carried out to x¹⁰, with the book's own n = 4 answer 1 − 4x² + (4/3)x⁴ dashed underneath it. Drag n: on every even integer the recurrence factor (2k − 2n) kills the tail and the curve snaps exactly onto a polynomial — at n = 4 it lands on the dashed reference and hides it — while at n = 3.5 nothing terminates and the two come apart.
      {"functions":\[{"color":"\#2e6b3f","domain":null,"expression":"1 - 4\*x^2 + (4/3)\*x^4","expression2":"","id":"herm-n4-ref","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#1f4e79","domain":null,"expression":"1 - n\*x^2 - n\*(2-n)/6\*x^4 - n\*(2-n)\*(4-n)/90\*x^6 - n\*(2-n)\*(4-n)\*(6-n)/2520\*x^8 - n\*(2-n)\*(4-n)\*(6-n)\*(8-n)/113400\*x^10","expression2":"","id":"herm-y1","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":1,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"herm-n","label":"Hermite order n","max":8,"min":0,"name":"n","step":0.5,"unit":"","value":4}\],"title":"Hermite equation: y1 series solution","version":1,"viewport":{"centerX":0,"centerY":0,"scale":85}}
    
    #examplebox("Example 3")[Hermite Equation][
Let us find a solution to the so-called Hermite’s equation of order #math.equation(block: false, alt: "n")[$n$]#math.equation(block: false, alt: "squared")[$2$] is the equation


        #math.equation(block: true, alt: "y double prime minus 2 x y prime plus 2 n y equals 0 .")[$y^(″) − 2 x y^(′) + 2 n y = 0 .$]


        Find a solution around the point #math.equation(block: false, alt: "x sub 0 equals 0")[$x_(0) = 0$].


      
      #solutionbox[
We try


        #math.equation(block: true, alt: "y equals ∑ k equals 0 infinity a sub k x to the power k .")[$y = ∑_(k = 0)^(∞) a_(k) x^(k) .$]


        We differentiate (as above) to obtain


        #math.equation(block: true, alt: "y prime equals ∑ k equals 1 infinity k a sub k x to the power k minus 1 ,; y double prime equals ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2 .")[$y^(′) = ∑_(k = 1)^(∞) k a_(k) x^(k − 1) , \ y^(″) = ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2) .$]


        Now we plug into the equation


        #math.equation(block: true, alt: "0 equals y double prime minus 2 x y prime plus 2 n y; equals open parenthesis ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2 close parenthesis minus 2 x open parenthesis ∑ k equals 1 infinity k a sub k x to the power k minus 1 close parenthesis plus 2 n open parenthesis ∑ k equals 0 infinity a sub k x to the power k close parenthesis; equals open parenthesis ∑ k equals 2 infinity k open parenthesis k minus 1 close parenthesis a sub k x to the power k minus 2 close parenthesis minus open parenthesis ∑ k equals 1 infinity 2 k a sub k x to the power k close parenthesis plus open parenthesis ∑ k equals 0 infinity 2 n a sub k x to the power k close parenthesis; equals open parenthesis 2 a sub 2 plus ∑ k equals 1 infinity open parenthesis k")[$0 = y^(″) − 2 x y^(′) + 2 n y \ = ( ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2) ) − 2 x ( ∑_(k = 1)^(∞) k a_(k) x^(k − 1) ) + 2 n ( ∑_(k = 0)^(∞) a_(k) x^(k) ) \ = ( ∑_(k = 2)^(∞) k ( k − 1 ) a_(k) x^(k − 2) ) − ( ∑_(k = 1)^(∞) 2 k a_(k) x^(k) ) + ( ∑_(k = 0)^(∞) 2 n a_(k) x^(k) ) \ = ( 2 a_(2) + ∑_(k = 1)^(∞) ( k + 2 ) ( k + 1 ) a_(k + 2) x^(k) ) − ( ∑_(k = 1)^(∞) 2 k a_(k) x^(k) ) + ( 2 n a_(0) + ∑_(k = 1)^(∞) 2 n a_(k) x^(k) ) \ = 2 a_(2) + 2 n a_(0) + ∑_(k = 1)^(∞) ( ( k + 2 ) ( k + 1 ) a_(k + 2) − 2 k a_(k) + 2 n a_(k) ) x^(k) .$]


        As #math.equation(block: false, alt: "y double prime minus 2 x y prime plus 2 n y equals 0")[$y^(″) − 2 x y^(′) + 2 n y = 0$] we have


        #math.equation(block: true, alt: "open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k plus 2 plus open parenthesis minus 2 k plus 2 n close parenthesis a sub k equals 0 , or a sub k plus 2 equals the fraction open parenthesis 2 k minus 2 n close parenthesis over open parenthesis k plus 2 close parenthesis open parenthesis k plus 1 close parenthesis a sub k .")[$( k + 2 ) ( k + 1 ) a_(k + 2) + ( − 2 k + 2 n ) a_(k) = 0 , #h(2em) "or" #h(2em) a_(k + 2) = frac(( 2 k − 2 n ), ( k + 2 ) ( k + 1 )) a_(k) .$]


        This recurrence relation actually includes #math.equation(block: false, alt: "a sub 2 equals minus n a sub 0")[$a_(2) = − n a_(0)$] (which comes about from #math.equation(block: false, alt: "2 a sub 2 plus 2 n a sub 0 equals 0")[$2 a_(2) + 2 n a_(0) = 0$]). Again #math.equation(block: false, alt: "a sub 0")[$a_(0)$] and #math.equation(block: false, alt: "a sub 1")[$a_(1)$] are arbitrary.


        #math.equation(block: true, alt: "a sub 2 equals the fraction minus 2 n over open parenthesis 2 close parenthesis open parenthesis 1 close parenthesis a sub 0 , a sub 3 equals the fraction 2 open parenthesis 1 minus n close parenthesis over open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis a sub 1 ,; a sub 4 equals the fraction 2 open parenthesis 2 minus n close parenthesis over open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis a sub 2 equals the fraction 2 squared open parenthesis 2 minus n close parenthesis open parenthesis minus n close parenthesis over open parenthesis 4 close parenthesis open parenthesis 3 close parenthesis open parenthesis 2 close parenthesis open parenthesis 1 close parenthesis a sub 0 ,; a sub 5 equals the fraction 2 open parenthesis 3 minus n close p")[$a_(2) = frac(− 2 n, ( 2 ) ( 1 )) a_(0) , #h(2em) a_(3) = frac(2 ( 1 − n ), ( 3 ) ( 2 )) a_(1) , \ a_(4) = frac(2 ( 2 − n ), ( 4 ) ( 3 )) a_(2) = frac(2^(2) ( 2 − n ) ( − n ), ( 4 ) ( 3 ) ( 2 ) ( 1 )) a_(0) , \ a_(5) = frac(2 ( 3 − n ), ( 5 ) ( 4 )) a_(3) = frac(2^(2) ( 3 − n ) ( 1 − n ), ( 5 ) ( 4 ) ( 3 ) ( 2 )) a_(1) , #h(1em) …$]


        Let us separate the even and odd coefficients. We find that


        #math.equation(block: true, alt: "a sub 2 m equals the fraction 2 to the power m open parenthesis minus n close parenthesis open parenthesis 2 minus n close parenthesis ⋯ open parenthesis 2 m minus 2 minus n close parenthesis over open parenthesis 2 m close parenthesis ! ,; a sub 2 m plus 1 equals the fraction 2 to the power m open parenthesis 1 minus n close parenthesis open parenthesis 3 minus n close parenthesis ⋯ open parenthesis 2 m minus 1 minus n close parenthesis over open parenthesis 2 m plus 1 close parenthesis ! .")[$a_(2 m) = frac(2^(m) ( − n ) ( 2 − n ) ⋯ ( 2 m − 2 − n ), ( 2 m ) !) , \ a_(2 m + 1) = frac(2^(m) ( 1 − n ) ( 3 − n ) ⋯ ( 2 m − 1 − n ), ( 2 m + 1 ) !) .$]


        Let us write down the two series, one with the even powers and one with the odd.


        #math.equation(block: true, alt: "y sub 1 open parenthesis x close parenthesis equals 1 plus the fraction 2 open parenthesis minus n close parenthesis over 2 ! x squared plus the fraction 2 squared open parenthesis minus n close parenthesis open parenthesis 2 minus n close parenthesis over 4 ! x to the power 4 plus the fraction 2 cubed open parenthesis minus n close parenthesis open parenthesis 2 minus n close parenthesis open parenthesis 4 minus n close parenthesis over 6 ! x to the power 6 plus ⋯ ,; y sub 2 open parenthesis x close parenthesis equals x plus the fraction 2 open parenthesis 1 minus n close parenthesis over 3 ! x cubed plus the fraction 2 squared open parenthesis 1 minus n close parenthesis open parenthesis 3 minus n close parenthesis over 5 ! x to the power 5 plus the fraction 2 cubed open parenthesis 1 mi")[$y_(1) ( x ) = 1 + frac(2 ( − n ), 2 !) x^(2) + frac(2^(2) ( − n ) ( 2 − n ), 4 !) x^(4) + frac(2^(3) ( − n ) ( 2 − n ) ( 4 − n ), 6 !) x^(6) + ⋯ , \ y_(2) ( x ) = x + frac(2 ( 1 − n ), 3 !) x^(3) + frac(2^(2) ( 1 − n ) ( 3 − n ), 5 !) x^(5) + frac(2^(3) ( 1 − n ) ( 3 − n ) ( 5 − n ), 7 !) x^(7) + ⋯ .$]


        We then write


        #math.equation(block: true, alt: "y open parenthesis x close parenthesis equals a sub 0 y sub 1 open parenthesis x close parenthesis plus a sub 1 y sub 2 open parenthesis x close parenthesis .")[$y ( x ) = a_(0) y_(1) ( x ) + a_(1) y_(2) ( x ) .$]


        We also notice that if #math.equation(block: false, alt: "n")[$n$] is a positive even integer, then #math.equation(block: false, alt: "y sub 1 open parenthesis x close parenthesis")[$y_(1) ( x )$] is a polynomial as all the coefficients in the series beyond a certain degree are zero. If #math.equation(block: false, alt: "n")[$n$] is a positive odd integer, then #math.equation(block: false, alt: "y sub 2 open parenthesis x close parenthesis")[$y_(2) ( x )$] is a polynomial. For example, if #math.equation(block: false, alt: "n equals 4")[$n = 4$], then


        #math.equation(block: true, alt: "y sub 1 open parenthesis x close parenthesis equals 1 plus the fraction 2 open parenthesis minus 4 close parenthesis over 2 ! x squared plus the fraction 2 squared open parenthesis minus 4 close parenthesis open parenthesis 2 minus 4 close parenthesis over 4 ! x to the power 4 equals 1 minus 4 x squared plus the fraction 4 over 3 x to the power 4 .")[$y_(1) ( x ) = 1 + frac(2 ( − 4 ), 2 !) x^(2) + frac(2^(2) ( − 4 ) ( 2 − 4 ), 4 !) x^(4) = 1 − 4 x^(2) + frac(4, 3) x^(4) .$]
]
]


    
=== Footnotes


    \[1\] Named after the English mathematician Sir George Biddell Airy (1801 – 1892).


    \[2\] Named after the French mathematician #link("http://en.Wikipedia.org/wiki/Hermite")[Charles Hermite] (1822–1901).