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Homework Help: Differential Equation w/ Power Series Solution

  1. Dec 12, 2004 #1

    [tex] y^{\prime} = x^2 y [/tex]

    General Comments

    There must be some kind of flaw in my solution as I don't get to the same result as the one my book provides:

    [tex] y = c_0 \sum _{n=0} ^{\infty} \frac{x^{3n}}{3^n n!} = c_0 e^{x^3 / 3} \qquad \fbox{CORRECT ANSWER} [/tex]

    Any help is highly appreciated.

    My Solution


    [tex] y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow x^2 y = \sum _{n=0} ^{\infty} c_n x^{n+2} = \sum _{n=3} ^{\infty} c_{n-3} x^{n-1} [/tex]


    [tex] y^{\prime} = \sum _{n=1} ^{\infty} n c_n x^{n-1} [/tex]


    [tex] \sum _{n=1} ^{\infty} n c_n x^{n-1} = \sum _{n=3} ^{\infty} c_{n-3} x^{n-1} \Longrightarrow c_{n-3} = nc_n \quad n=3,4,5,\ldots [/tex]

    Hence, I ultimately get

    [tex] y = c_0 + c_1 x + c_2 x^2 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] + c_1 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-2}}{4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n-2 \right) } \right] + c_2 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-1}}{5\cdot 8\cdot 11\cdot \cdots \cdot \left( 3n-1 \right) } \right] \qquad \fbox{MY ANSWER} [/tex]

    Thank you very much!!
  2. jcsd
  3. Dec 12, 2004 #2


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    If you do your recursion relation carefully you will find that [itex]c_1 = c_2 = 0[/itex] since [itex]c_n = 0[/itex] for n < 0.
  4. Dec 12, 2004 #3
    Oh... I've got it. Thanks
  5. Dec 12, 2004 #4


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    Cheating method:Why bother with stinky recursion relations,when u can use a series already known?? :wink:

    [tex]\frac{dy}{dx} =y x^2 [/tex] from which,by separating variables and integrating,one gets:
    [tex]y=C\exp(\frac{x^3}{3}) [/tex].To find the series the problem asks you about,plug [itex] \frac{x^3}{3} [/itex] in the power series of [itex]\exp{x} [/itex] and you'll have your answer.

    PS.This cheating trick works,as the differential eq.is easily solvable exactly.In general,it does not. :wink:
  6. Dec 12, 2004 #5
    Just a final check...


    Method #1

    How about...

    [tex] y = c_0 + c_1 x + c_2 x^2 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] + c_1 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-2}}{4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n-2 \right) } \right] + c_2 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-1}}{5\cdot 8\cdot 11\cdot \cdots \cdot \left( 3n-1 \right) } \right] [/tex]

    So, [tex] c_1 = c_2 = 0 [/tex] ( I don't see clearly why ) . Then, we find

    [tex] y = c_0 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] = c_0 \sum _{n=0} ^{\infty} \frac{x^{3n}}{3^{n}n!} = c_0 e^{x^3 /3} [/tex]


    Method #2

    [tex] \frac{dy}{dx} = x^2 y \Longrightarrow \int \frac{1}{y}dy = \int x^2 dx [/tex]


    [tex] \ln y = \frac{x^3}{3} \Longrightarrow y = Ce^{x^3/3} [/tex]

    where [tex] C = c_0 [/tex]
    Last edited: Dec 12, 2004
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