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anonymous

  • one year ago

Y"+x^2y'+xy=0 power series

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  1. anonymous
    • one year ago
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    About \(x=0\)?

  2. anonymous
    • one year ago
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    Yes

  3. Jacob902
    • one year ago
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    y = ∑(n=0 to ∞) a_n x^n y' = ∑(n=0 to ∞) a_n n x^(n-1) y'' = ∑(n=0 to ∞) a_n n(n-1) x^(n-2) y'' - x^2 y' - xy = = ∑(n=0 to ∞) a_n n(n-1) x^(n-2) - - ∑(n=0 to ∞) a_n n x^(n+1) - - ∑(n=0 to ∞) a_n x^(n+1) = = ∑(n=-1 to ∞) a_(n+3) (n+3)(n+2) x^(n+1) - - ∑(n=0 to ∞) a_n n x^(n+1) - - ∑(n=0 to ∞) a_n x^(n+1) = 0 The first sum starts at n=-1, the other two start at n=0. Then, the term n=-1 of the first sum must be zero: a_2 ∙ 2 ∙ 1 = 0 => a_2 = 0 For n≥0: a_(n+3) (n+3)(n+2) - a_n (n+1) = 0 => a_(n+3) = a_n (n+1) / [(n+3)(n+2)] It's right: you can choose a_0 and a_1 arbitrarily (order 2, 2 initial conditions to adjust), and a_2 must be zero. All other a_n are given by the previous recursion.

  4. anonymous
    • one year ago
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    \[\begin{align*} y&=\sum_{n=0}^\infty a_nx^n\\[1ex] y'&=\sum_{n=1}^\infty na_nx^{n-1}\\[1ex] y''&=\sum_{n=2}^\infty n(n-1)a_nx^{n-2} \end{align*}\] Substituting into the ODE: \[\begin{align*} 0&=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+x^2\sum_{n=1}^\infty na_nx^{n-1}+x\sum_{n=0}^\infty a_nx^n\\[2ex] &=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n+1}+\sum_{n=0}^\infty a_nx^{n+1}\\[2ex] &=\sum_{n+3=2}^\infty (n+3)(n+3-1)a_{n+3}x^{n+3-2}+\sum_{n=1}^\infty na_nx^{n+1}+\sum_{n=0}^\infty a_nx^{n+1}\\[2ex] &=\sum_{n=-1}^\infty (n+3)(n+2)a_{n+3}x^{n+1}+\sum_{n=1}^\infty na_nx^{n+1}+\sum_{n=0}^\infty a_nx^{n+1}\\[2ex] &=\left(2a_2+6a_3x+\sum_{n=1}^\infty (n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n=1}^\infty na_nx^{n+1}\\[1ex] &\quad\quad+\left(a_0x+\sum_{n=1}^\infty a_nx^{n+1}\right)\\[2ex] &=2a_2+(6a_3+a_0)x+\sum_{n=1}^\infty \bigg[(n+3)(n+2)a_{n+3}+(n+1)a_n\bigg]x^{n+1} \end{align*}\]

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