Question 8.4.1: Find the solution of the initial-value problem (1−t^2)y″−2t...

First, observe that the given DE is Legendre’s equation with λ = 3, since 3(3 + 1) = 12. From our earlier work in this section, we know that the general solution is

y(t ) = c_{1}P_{3}(t )+c_{2}Q_{3}(t )

= c_{1}P_{3}(t )+c_{2}(P_{3}(t )Q_{0}(t )− \frac {5}{2}t^{2} + \frac {2}{3})

= P_{3}(t ))(c_{1} +c_{2}Q_{0}(t ))+c_{2}( -\frac {5}{2}t^{2} + \frac {2}{3})

=(\frac {5}{2}t^{3} − \frac {3}{2}t) (c_{1} + \frac {c_{2}}{2} ln \frac {1−t}{1+t}) +c_{2}(−\frac {5}{2}t^{2} + \frac {2}{3}            (8.4.14)

Applying the initial conditions y(0) = 1 and y′(0) = 1 to 8.4.14, we can
show that c_{1} =−2/3 and c_{2} = 3/2, and thus

y=(\frac {5}{2}t^{3} − \frac {3}{2}t)(-\frac {2}{3} + \frac {3}{4} ln \frac {1−t}{1+t})− \frac {15}{4}t^{2} +1

is the solution to the given IVP.