Word Count : 450

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हमें निम्नलिखित अवकल समीकरण का हल फोबेनियस विधि से निकालना है: [ \frac{d^2y}{dx^2} + \frac{1}{2x}\frac{dy}{dx} + y = 0 ] यहाँ (x=0) पर एक नियमित संधि (regular singular point) है, इसलिए हम फोबेनियस विधि का उपयोग करेंगे। फोबेनियस विधि के अनुसार हम हल मानते हैं: [ y = x^r \sum_{n=0}^{\infty} a_n x^n, \quad a_0 \neq 0 ] पहले अवकलन निकालते हैं: [ \frac{dy}{dx} = \sum_{n=0}^{\infty} a_n (n+r) x^{n+r-1}, \quad \frac{d^2y}{dx^2} = \sum_{n=0}^{\infty} a_n (n+r)(n+r-1) x^{n+r-2} __________ ____ ________ _____ _____ ________.
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