--- title: "Series solution of x²y''+xy'+(x²−4)y=0" subject: "Mathematics" chapter: 4 key: "semester_2/mathematics:series-solution-x2y-xy-x2" source: https://www.bctengineer.com/q/semester_2/mathematics/series-solution-x2y-xy-x2 site: BCT Engineering --- # Series solution of x²y''+xy'+(x²−4)y=0 Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions - Asked in: 2069 Poush, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/series-solution-x2y-xy-x2 - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4 ## Question Solve the equation x² d²y/dx² + x dy/dx + (x²−4)y = 0 in series form. ## Short answer This is Bessel's equation of order 2. The Frobenius method gives the series J₂(x)=Σ(-1)^k/(k!(k+2)!)(x/2)^(2k+2) and a logarithmic second solution Y₂(x). General solution: y=A J₂(x)+B Y₂(x). ## Answer Equation: x²y″ + xy′ + (x² − 4)y = 0. This is Bessel's equation of order 2. x = 0 is a regular singular point, so use the Frobenius method: y = Σ aₖ x^(k+m), a₀ ≠ 0. #### Indicial equation and recurrence Substituting, the coefficient of x^(k+m) gives [(k+m)² − 4] aₖ + aₖ₋₂ = 0. - Lowest power (k = 0): m² − 4 = 0, so m = 2 or −2. - Recurrence: aₖ = − aₖ₋₂ / ((k + m)² − 4), and a₁ = a₃ = … = 0. #### Root m = 2 aₖ = − aₖ₋₂ / (k(k + 4)): ```text a₂ = −a₀/12, a₄ = a₀/(12·32) = a₀/384, a₆ = −a₀/(384·60) … ``` Taking a₀ = 1/8 (i.e. 1/(2²·2!)): ```text y₁ = J₂(x) = x²/8 − x⁴/96 + x⁶/3072 − … = Σ (−1)ᵏ / (k!(k+2)!) · (x/2)^(2k+2) ``` #### Root m = −2 The roots differ by an integer (4). With m = −2 the recurrence is aₖ = −aₖ₋₂/(k(k − 4)), which fails at k = 4 (division by zero, while a₂ = a₀/4 ≠ 0). So the second solution contains a logarithm and is not a pure power series; it is the Bessel function of the second kind Y₂(x) (J₋₂ = J₂ for integer order, so it is not independent). **Answer: y = A J₂(x) + B Y₂(x)**, where J₂(x) = Σ (−1)ᵏ/(k!(k+2)!)(x/2)^(2k+2) is the series solution above. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. x^2y'' + xy' + (x^2 − 4)y = 0 has x = 0 as what, so which method is used? A. An ordinary point; Taylor series B. An irregular point; no series exists C. An ordinary point; Frobenius method D. A regular singular point; Frobenius method Answer: D. The coefficient of y'' vanishes at 0, but it is regular singular, so Frobenius' y = Σ aₖ x^(k+m) is used. (Key point: Method and point) 2. The indicial equation and its roots are: A. m^2 + 4 = 0, m = ±2i B. m^2 − 2 = 0, m = ±√2 C. m^2 − 4 = 0, m = ±2 D. m − 4 = 0, m = 4 Answer: C. The lowest power gives m^2 − 4 = 0. (Key point: Indicial equation) 3. Why can the root m = −2 not give a second pure power series? A. a0 must be zero B. The roots differ by an integer and the recurrence divides by zero at k = 4 C. J₋₂ is independent of J₂ but diverges D. The roots are complex Answer: B. With m = −2 the recurrence has k(k − 4), which fails at k = 4; a log term appears (Y₂). (Key point: Second solution) 4. The general solution is: A. y = A J₂(x) + B Y₂(x) B. y = A J₂(x) + B J₋₂(x) C. y = A J₂(x) only D. y = A cos 2x + B sin 2x Answer: A. For integer order J₋₂ = J₂, so the second independent solution is Y₂. 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