Question

(x^3 - 1)y''(x) + (x^2)y'(x) + xy(x) = 0, y(0)=2, y'(0)=1

Answer 1

Answer on Question #82282 – Math – Differential Equations

Question

$$(x^3 - 1) y''(x) + (x^2) y'(x) + x y(x) = 0, y(0) = 2, y'(0) = 1$$

Solution

$$y = \sum_{n=0}^{\infty} a_n x^n$$ $$y' = \sum_{n=1}^{\infty} a_n n x^{n-1}$$ $$y'' = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}$$ $$(x^3 - 1) \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} + (x^2) \sum_{n=1}^{\infty} a_n n x^{n-1} + x \sum_{n=0}^{\infty} a_n x^n = 0$$ $$\sum_{n=2}^{\infty} a_n n(n-1) x^{n+1} - \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} + \sum_{n=1}^{\infty} a_n n x^{n+1} + \sum_{n=0}^{\infty} a_n x^{n+1} = 0$$ $$n - 2 \to N, n \to N + 2$$ $$\sum_{n=2}^{\infty} a_n n(n-1) x^{n+1} - \sum_{N+2=2}^{\infty} a_{N+2} (N+2) (N+2-1) x^{N+2-2} +$$ $$+ \sum_{n=1}^{\infty} a_n n x^{n+1} + \sum_{n=0}^{\infty} a_n x^{n+1} = 0$$ $$n + 1 \to N, n \to N - 1$$ $$\sum_{N-1=2}^{\infty} a_{N-1} (N-1) (N-1-1) x^N - \sum_{N=0}^{\infty} a_{N+2} (N+2) (N+2-1) x^N +$$ $$+ \sum_{N-1=1}^{\infty} a_{N-1} (N-1) x^N + \sum_{N-1=0}^{\infty} a_{N-1} x^N = 0$$ $$\sum_{N=3}^{\infty} a_{N-1} (N-1) (N-1-1) x^N - a_2(2)(1) - a_3(3)(2) x - a_4(4)(3) x^2 -$$ $$- \sum_{N=3}^{\infty} a_{N+2}(N+2)(N+1)x^{N} + a_{1}(1)x^{2} + \sum_{N=3}^{\infty} a_{N-1}(N-1)x^{N} + a_{0}x +$$ $$+ a_{1}x^{2} + \sum_{N=3}^{\infty} a_{N-1}x^{N} = 0$$ $$\sum_{N=3}^{\infty} \left(a_{N-1}\left(N^{2} - 3N + 2 + N - 1 + 1\right) - a_{N+2}\left(N^{2} + 3N + 2\right)\right)x^{N} - 2a_{2} -$$ $$- 6a_{3}x - 12a_{4}x^{2} + a_{1}x^{2} + a_{0}x + a_{1}x^{2} = 0$$ $$- 2a_{2} = 0$$ $$a_{0} - 6a_{3} = 0$$ $$- 12a_{4} + 2a_{1} = 0$$ $$a_{N-1}(N^{2} - 2N + 2) - a_{N+2}(N^{2} + 3N + 2) = 0$$ $$y = \sum_{n=0}^{\infty} a_{n}x^{n}$$ $$y(0) = 2$$ $$a_{0} = 2$$ $$y' = \sum_{n=1}^{\infty} a_{n}nx^{n-1}$$ $$y'(0) = 1$$ $$a_{1} = 1$$ $$a_{2} = 0$$ $$a_{3} = \frac{1}{3}$$ $$a_{4} = \frac{1}{6}$$ $$a_{N+2} = \frac{N^{2} - 2N + 2}{N^{2} + 3N + 2} \cdot a_{N-1}$$ $$y = 2 + x + \sum_{n=1}^{\infty} a_{n+2}x^{n+2}$$ $$a_{n+2} = \frac{n^{2} - 2n + 2}{n^{2} + 3n + 2} \cdot a_{n-1}, \quad a_{0} = 2.$$

Answer: $y = 2 + x + \sum_{n=1}^{\infty} a_{n+2}x^{n+2}$, $a_{n+2} = \frac{n^{2} - 2n + 2}{n^{2} + 3n + 2} \cdot a_{n-1}$, $a_{0} = 2$.

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