JNTU B.Tech II MATHEMATICS-III Previous Question Paper (October - 2020 - R15)

The given differential equation is (1 - x²)y'' - 2xy' + 2y = 0. Since x = 0 is an ordinary point, we assume a power series solution of the form:

Differentiating term by term:

Substituting these into the differential equation:

Expanding and simplifying the summation indices:

Shifting the index of the first summation by setting k = n-2, we get ∑_k=0^∞ (k+2)(k+1)a_k+2x^k. Combining all terms with xⁿ:

Setting the coefficients to zero:

1. 2a₂ + 2a₀ = 0 ⇒ a₂ = -a₀
2. 6a₃ = 0 ⇒ a₃ = 0
3. Recurrence relation: a_n+2 = [(n²+n-2) / ((n+2)(n+1))]a_n = [(n+2)(n-1) / ((n+2)(n+1))]a_n = [(n-1)/(n+1)]a_n

Calculating further coefficients:

Since all odd terms a₃, a₅, ... are zero, the series solution is:

The given differential equation is x y'' + 2 y' + 3 y = 0. Since the coefficient of y'' vanishes at x = 0, we check for a singularity. Dividing by x, we get y'' + (2/x) y' + (3/x) y = 0. The point x = 0 is a regular singular point because x(2/x) = 2 and x²(3/x) = 3x are analytic at x = 0. We use the Frobenius method, assuming a solution of the form y = Σ a_n x^n+r where a₀ ≠ 0.

Substituting y = Σ a_n x^n+r, y' = Σ (n+r) a_n x^n+r-1, and y'' = Σ (n+r)(n+r-1) a_n x^n+r-2 into the equation:

Combining terms with x^n+r-1, the indicial equation is obtained by setting n=0: r(r-1) + 2r = 0, which simplifies to r(r+1) = 0. Thus, the roots are r₁ = 0 and r₂ = -1.

For the recurrence relation, equating the coefficient of x^n+r-1 to zero:

Case 1: r = 0. The relation is a_n = -3 a_n-1 / [n(n+1)]. For n=1, a₁ = -3a₀/2. For n=2, a₂ = -3a₁/6 = a₀/4. The general term is a_n = (-3)ⁿ a₀ / [n!(n+1)!]. The first solution is y₁ = a₀ Σ [(-3)ⁿ xⁿ / (n!(n+1)!)].

Case 2: r = -1. The relation is a_n = -3 a_n-1 / [n(n-1)]. This leads to a singularity at n=1, requiring the inclusion of a logarithmic term or checking for the Frobenius series existence. The general solution is y = c₁ y₁ + c₂ y₂, where y₂ involves the second root and potentially a log term.