Question
Solve the following ODE using the Frobenius method:
Answer :
Word Count : 552
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Skip in 10
We are asked to solve the differential equation [ \frac{d^2y}{dx^2} + \frac{1}{2x} \frac{dy}{dx} + y = 0 ] using the Frobenius method. Let's solve it step by step manually. --- Assume a solution of the form [ y = x^r \sum_{n=0}^{\infty} a_n x^n, \quad a_0 \neq 0 ] so that [ y = \sum_{n=0}^{\infty} a_n x^{n+r}. ] Then [ y' = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1}, \quad y'' = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2}. ] Substitute into the ODE: [ \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} + \frac{1}{2x} \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} + \sum_{n=0}^{\infty} a_n x^{n+r} = 0. ] Simplify the second term: [ \frac{1}{2x} \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1} = \sum_{n=0}^{\infty} \frac{1}{2} a_n (n+r) x^{n+r-2+1} = \sum_{n=0}^{\infty} \frac{1}{2} a_n (n+r) x^{n+r-2+1} = \sum_{n=0}^{\infty} \frac{1}{2} a_n
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आवर्तकाल वाले निम्नलिखित आवर्ती फलन का फूरिए श्रेणी प्रसार प्राप्त करें :
के लिए।
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