Question 8.3.3: Consider the linear second-order differential equation y″−2...

We begin by assuming that y = a_{0} +a_{1}t +a_{2}t^{2} +a_{3}t^{3}+· · · . From this, it follows

y′= a_{1} +2a_{2}t +3a_{3}t^{2} +4a_{4}t^{3}+· · · =\sum\limits_{n=1}^{∞}{na_{n}t^{n−1}}

−2ty′=−2at −4a_{2}t^{2} −6a_{3}t^{3} −8a_{4}t^{3+}· · ·=-\sum\limits_{n=1}^{∞}{2na_{n}t^{n}}

y″=2a_{2} +6a_{3}t +12a_{4}t^{2} +20a_{5}t^{3}+· · ·=\sum\limits_{n=2}^{∞}{n(n−1)a_{n}t^{n−2}}

In many instances, it will be most convenient to work with power series represented in the shorthand sigma (∑) notation, which is how we will proceed from here. Substituting in (8.3.11) with the series expressions for y″, −2ty, and y′, we find

\sum\limits_{n=2}^{∞}{n(n−1)a_{n}t^{n−2}}-\sum\limits_{n=1}^{∞}{2na_{n}t^{n}}+\sum\limits_{n=0}^{∞}{a_{n}t^{n}}=0                   (8.3.12)

In order to equate the coefficients of like powers of t , it is helpful to write each series in (8.3.12) using the same indices for the sum. Replacing n with n + 2 allows us to write

\sum\limits_{n=2}^{∞}{n(n−1)a_{n}t^{n−2}}\sum\limits_{n=0}^{∞}{(n+2)(n+1)a_{n+2}t^{n}}

In addition, observe that

\sum\limits_{n=1}^{∞}{2na_{n}t^{n}}=\sum\limits_{n=0}^{∞}{2na_{n}t^{n}}

because the term −2na_{n} vanishes when n = 0. Therefore we can revise (8.3.12) to have the form

\sum\limits_{n=0}^{∞}{(n+2)(n+1)a_{n+2}t^{n}}+\sum\limits_{n=0}^{∞}{−2na_{n}t^{n}}+\sum\limits_{n=0}^{∞}{a_{n}t^{n}}=0  (8.3.13)

Now that each series is indexed from n = 0 with corresponding powers of t, we can combine the three sums into one and write

\sum\limits_{n=0}^{∞}{[(n+2)(n+1)a_{n+2} −2na_{n} +a_{n]}t^{n}}                     (8.3.14)

Because (8.3.14) implies that every coefficient of the series must be zero, we see that the constants an must satisfy the recurrence relation

(n+2)(n+1)a_{n+2} −2na_{n} +a_{n} = 0

or equivalently

a{n+2} = \frac{2n−1}{(n+2)(n+1)}a_{n}, n = 0,1,2, . . . .            (8.3.15)

Here it is essential to observe that since the subscripts differ by two in (8.3.15), we can obtain two distinct series solutions to the original equation (8.3.11), one involving all of the even terms and the other all of the odd ones. In particular, considering n = 0,2,4, . . ., we have from (8.3.15) that

a_{2} =−\frac {1}{2}a_{0}, a_{4} = \frac {3}{3 · 4}a_{2} =\frac {−1 · 3}{2 · 3 · 4}a_{0},  and  a_{6} = \frac {7}{6 · 5}a_{4} =\frac { −1 · 3 · 7}{2 · 3 · 4 · 5 · 6}a_{0}

More generally, the pattern

a_{2n} \frac {−1 · 3 · 7· · · (4n−5)}{(2n)!}

holds and therefore

y_{1}(t ) = a_{0} − \frac {1}{2}a_{0}t^{2} − \frac {1}{8}a_{0}t^{4} − \frac {7}{240}a_{0}t^{6}+· · ·

=a_{0} −a_{0} \sum\limits_{n=1}^{∞}{\frac {1 · 3 · 7· · · (4n−5)}{(2n)!} t^{2n}}                     (8.3.16)

Similarly, if we examine the odd terms for n = 1,3,5, . . . in (8.3.15), we see

a_{3} = \frac {1}{2 · 3}a_{1}, a_{5} = \frac {5}{4 · 5}a_{3} = \frac {1 · 5}{2 · 3 · 4 · 5}a1,   and   a_{7} = \frac {9}{6 · 7}a_{5} = \frac {1 · 5 · 9}{2 · 3 · 4 · 5 · 6 · 7}a_{1}

Thus, we find

a_{2n+1} = \frac {1 · 5 · 9· · · (4n−3)}{(2n+1)!} a_{1}

and therefore

y_{2}(t ) = a_{1}t + \frac {1}{6}a_{1}t^{3} + \frac {1}{24}a_{1}t^{5}+· · ·

= a_{1}t +a_{1}\sum\limits_{n=1}^{∞}{\frac {1· 5 · 9· · · (4n−3)}{(2n+1)!} t^{2n+1}}                     (8.3.17)

Because y_{1} only involves even powers of t and y_{2} only involves odd powers of t , it is obvious that y_{1} and y_{2} must be linearly independent functions: it is impossible for one to be a scalar multiple of the other. Hence we have found the two basic solutions to the given DE and the general solution is

y = a_{0}y_{1}+a_{1}y_{2}

y= a_{0}(1 − \sum\limits_{n=1}^{∞}{\frac {1 · 3 · 7· · · (4n−5)}{(2n)!} t^{2n}})+a_{1}(t +\sum\limits_{n=1}^{∞}{\frac {1· 5 · 9· · · (4n−3)}{(2n+1)!} t^{2n+1}})

Moreover, since p(t ) = −2t and q(t ) = 1 are analytic everywhere, it follows from theorem 8.3.1 that both y_{1} and y_{2} converge for all values of t , as must the general solution (8.3.18).

Finally, if we desire to solve the initial-value problem with y(0) = 2 and y′(0) = −1, we need only observe from our beginning assumption regarding the series expansion of y that y(0)=a_{0} =2 and y′(0)=a_{1}=−1. Therefore, the solution to the IVP is

y= 2(1 − \sum\limits_{n=1}^{∞}{\frac {1 · 3 · 7· · · (4n−5)}{(2n)!} t^{2n}})-(t +\sum\limits_{n=1}^{∞}{\frac {1· 5 · 9· · · (4n−3)}{(2n+1)!} t^{2n+1}})