- #1
DivGradCurl
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Any help is highly appreciated.
Thank you
Problem
[tex] y^{\prime \prime} = xy [/tex]
My Solution
If
[tex] y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow xy = \sum _{n=0} ^{\infty} c_n x^{n+1} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]
and
[tex] y^{\prime \prime} = \sum _{n=2} ^{\infty} \left( n-1 \right) n c_n x^{n-2} = \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} [/tex]
Then
[tex] \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]
Hence, we may find the recursion relation
[tex] c_{n+2} = \frac{c_{n-1}}{\left( n+1 \right) \left( n+2 \right)} \qquad n=1,2,3,\ldots [/tex]
and so we have:
[tex] c_0 [/tex]
[tex] c_1 [/tex]
[tex] c_2 [/tex]
[tex] c_3 = \frac{c_0}{2\cdot 3} [/tex]
[tex] c_4 = \frac{c_1}{3\cdot 4} [/tex]
[tex] c_5 = \frac{c_2}{4\cdot 5} [/tex]
[tex] c_6 = \frac{c_3}{5\cdot 6} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6} [/tex]
[tex] c_7 = \frac{c_4}{6\cdot 7} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7} [/tex]
[tex] c_8 = \frac{c_5}{7\cdot 8} = \frac{c_2}{4\cdot 5\cdot 7\cdot 8} [/tex]
[tex] \vdots [/tex]
Well, since this is a second-degree differential equation, we must have [tex] c_2 = 0 [/tex]. It also follows that
[tex] c_{3n} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \qquad n=1,2,3,\ldots [/tex]
and
[tex] c_{3n+1} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \qquad n=1,2,3,\ldots [/tex]
Therefore
[tex] y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right] [/tex]
Questions
1. Have I got it right?
2. If so, can I write the denominators in a more elegant way? I mean:
[tex] 2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 2\cdot 5\cdot 8\cdot \cdots \cdot \left( 3n-1 \right) \right] = n!3^n \times ( ? ) [/tex]
and
[tex] 3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n+1 \right) \right] = n!3^n \times ( ? ) [/tex]
where [tex] ( ? ) [/tex] represents the other factor which I haven't been able to figure out so far.
Thank you
Problem
[tex] y^{\prime \prime} = xy [/tex]
My Solution
If
[tex] y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow xy = \sum _{n=0} ^{\infty} c_n x^{n+1} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]
and
[tex] y^{\prime \prime} = \sum _{n=2} ^{\infty} \left( n-1 \right) n c_n x^{n-2} = \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} [/tex]
Then
[tex] \sum _{n=0} ^{\infty} \left( n+1 \right) \left( n+2 \right) c_{n+2} x^{n} = \sum _{n=1} ^{\infty} c_{n-1} x^n [/tex]
Hence, we may find the recursion relation
[tex] c_{n+2} = \frac{c_{n-1}}{\left( n+1 \right) \left( n+2 \right)} \qquad n=1,2,3,\ldots [/tex]
and so we have:
[tex] c_0 [/tex]
[tex] c_1 [/tex]
[tex] c_2 [/tex]
[tex] c_3 = \frac{c_0}{2\cdot 3} [/tex]
[tex] c_4 = \frac{c_1}{3\cdot 4} [/tex]
[tex] c_5 = \frac{c_2}{4\cdot 5} [/tex]
[tex] c_6 = \frac{c_3}{5\cdot 6} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6} [/tex]
[tex] c_7 = \frac{c_4}{6\cdot 7} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7} [/tex]
[tex] c_8 = \frac{c_5}{7\cdot 8} = \frac{c_2}{4\cdot 5\cdot 7\cdot 8} [/tex]
[tex] \vdots [/tex]
Well, since this is a second-degree differential equation, we must have [tex] c_2 = 0 [/tex]. It also follows that
[tex] c_{3n} = \frac{c_0}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \qquad n=1,2,3,\ldots [/tex]
and
[tex] c_{3n+1} = \frac{c_1}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \qquad n=1,2,3,\ldots [/tex]
Therefore
[tex] y = c_0 + c_1 x + c_0 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n}}{2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right)} \right] + c_1 \sum_{n=1} ^{\infty} \left[ \frac{x^{3n+1}}{3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right)} \right] [/tex]
Questions
1. Have I got it right?
2. If so, can I write the denominators in a more elegant way? I mean:
[tex] 2\cdot 3\cdot 5\cdot 6\cdot \cdots \cdot \left( 3n -1 \right) \left( 3n \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 2\cdot 5\cdot 8\cdot \cdots \cdot \left( 3n-1 \right) \right] = n!3^n \times ( ? ) [/tex]
and
[tex] 3\cdot 4\cdot 6\cdot 7\cdot \cdots \cdot \left( 3n \right) \left( 3n +1 \right) = \left[ 3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) \right] \left[ 4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n+1 \right) \right] = n!3^n \times ( ? ) [/tex]
where [tex] ( ? ) [/tex] represents the other factor which I haven't been able to figure out so far.