JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II MATHEMATICS-III Previous Question Paper (March - 2022 - R13)

This is a Cauchy-Euler equation. Let t = e^z, which implies z = ln(t). The derivatives are given by:

where D = d/dz. Substituting these into the equation:

The characteristic equation is m^2 + 6m + 5 = 0, which factors to (m+1)(m+5) = 0, giving roots m = -1, -5. The complementary function is:

For the particular integral y_p, let y_p = A e^z. Substituting into the transformed equation:

Thus, y_p = (1/12)e^z = t/12. The general solution is:

Assume a power series solution of the form y = Σ a_n x^n. Then y' = Σ n a_n x^{n-1} and y'' = Σ n(n-1) a_n x^{n-2}. Substituting into y'' - xy' + y = 0:

Shifting indices for the first term:

For n=0: 2(1)a_2 + a_0 = 0 => a_2 = -a_0/2. For n >= 1:

Calculating coefficients:

• For n=1: a_3 = 0, so all odd terms a_3, a_5, ... are 0.
• For n=2: a_4 = (1/12)a_2 = -a_0/24.
• For n=4: a_6 = (3/30)a_4 = (1/10)(-a_0/24) = -a_0/240.

The solution is y = a_0(1 - x^2/2 - x^4/24 - x^6/240 - ...) + a_1 x.