JNTU B.Tech II MATHEMATICS-III Previous Question Paper (April/May - 2018 - R15)

The given equation is a Cauchy-Euler equation. Let x = e^t, then x(dy/dx) = Dy and x²(d²y/dx²) = D(D-1)y, where D = d/dt. Substituting these, the equation becomes [D(D-1) - 6D + 10]y = e^2t, which simplifies to (D² - 7D + 10)y = e^2t. The particular integral is PI = 1/(D² - 7D + 10) * e^2t. Since substituting D=2 gives 4 - 14 + 10 = 0, we use the formula 1/f(D) * e^at = t * e^at / f'(a). Thus, PI = t * e^2t / (2D - 7) evaluated at D=2, which is t * e^2t / (4 - 7) = -1/3 * t * e^2t. Substituting back t = ln(x), the PI is -1/3 * x² ln(x).

The equation is (d²y/dx²) + [2(x-1) / (x³(x-1))] (dy/dx) + [1 / (x³(x-1))] y = 0. Simplifying, P(x) = 2/x³ and Q(x) = 1/(x³(x-1)). The singular points are where the coefficients are not analytic, namely x = 0 and x = 1. For x = 1, lim_x→1 (x-1)P(x) = 0 and lim_x→1 (x-1)²Q(x) = 0, so x = 1 is a regular singular point. For x = 0, lim_x→0 xP(x) = lim_x→0 2/x² = ∞, so x = 0 is an irregular singular point.