JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MATHEMATICS-III Previous Question Paper (September - 2021 - R15)

This is a Cauchy-Euler differential equation. Let x = e^t, which implies t = log x. The derivatives are transformed as follows: x(dy/dx) = Dy, x²(d²y/dx²) = D(D-1)y, and x³(d³y/dx³) = D(D-1)(D-2)y, where D = d/dt.

Substituting these into the equation: [D(D-1)(D-2) + 3D(D-1) + D + 1]y = (e^t)² t. Simplifying the operator: [D³ - 3D² + 2D + 3D² - 3D + D + 1]y = t e^2t, which results in (D³ + 1)y = t e^2t.

The auxiliary equation is m³ + 1 = 0, giving roots m = -1 and m = (1 ± i√3)/2. The complementary function is y_c = c₁e^-t + e^t/2(c₂ cos(√3t/2) + c₃ sin(√3t/2)).

For the particular integral y_p = 1/(D³ + 1) * (t e^2t) = e^2t * 1/((D+2)³ + 1) * t = e^2t * 1/(D³ + 6D² + 12D + 9) * t. Using binomial expansion: e^2t * (1/9)(1 + (12D + 6D² + D³)/9)^-1 * t = (e^2t/9)(1 - 4D/3)t = (e^2t/9)(t - 4/3).

Substituting back x = e^t, the general solution is y = c₁/x + √x(c₂ cos(√3 log x / 2) + c₃ sin(√3 log x / 2)) + (x²/9)(log x - 4/3).

This is a Legendre's linear differential equation. Let 1+x = e^t, so t = log(1+x). The substitutions are (1+x)(dy/dx) = Dy and (1+x)²(d²y/dx²) = D(D-1)y, where D = d/dt.

Substituting into the equation: [D(D-1) + D + 1]y = 2 sin t, which simplifies to (D² + 1)y = 2 sin t.

The auxiliary equation is m² + 1 = 0, giving roots m = ±i. The complementary function is y_c = c₁ cos t + c₂ sin t.

For the particular integral y_p = 1/(D² + 1) * 2 sin t. Since the denominator becomes zero at D² = -1, we use the formula 1/(D² + a²) * sin(at) = -t/(2a) * cos(at). Thus, y_p = -t/(2*1) * 2 cos t = -t cos t.

Substituting back t = log(1+x), the general solution is y = c₁ cos(log(1+x)) + c₂ sin(log(1+x)) - log(1+x) cos(log(1+x)).