Question 8.3.2: Solve the initial-value problem y″+y = 0, y(0) = 1, y′(0) =...

Since y=a_{0}+a_{1}t +a_{2}t^{2}+a_{3}t^{3}+a_{4}t^{4}+· · , it follows that

y′= a_{1} +2a_{2}t +3a_{3}t^{2} +4a_{4}t^{3}+· · ·    and
y″= 2a_{2} +3 · 2a_{3}t +4 · 3a_{4}t^{2} +5 · 4a_{5}t^{3}+· · ·

stituting for y and y″ in the given equation y″+y = 0, we have

(a_{0}+a_{1}t +a_{2}t^{2}+a_{3}t^{3}+a_{4}t^{4}+· ·)+(2a_{2} +3 · 2a_{3}t +4 · 3a_{4}t^{2} +5 · 4a_{5}t^{3}+· · ·)=0

Gathering terms with like coefficients,

(a_{0} +2a_{2})+(a_{1} +6a_{3})t +(a_{2} +12a_{4})t^{2} +(a_{3} +20a_{5})t^{3}+· · ·=0             (8.3.4)

etting each coefficient of powers of t in (8.3.4) equal to zero implies that the following sequence of equalities holds:

a_{0} =−2a_{2}     a_{1} =−6a_{3}
a_{2} =−12a_{4}      a_{3} =−20a_{5}
a_{4} =−30a_{6}     a_{5} =−42a_{7}
¦                      ¦
a_{2n} =−(2n+2)(2n+1)a_{2n+2}      a_{2n+1} =−(2n+3)(2n+2)a_{2n+3}

we group these equations into the two columns shown for the natural reason that the coefficients with even indices depend recursively on one another, as do the coefficients with odd indices. Furthermore, we see that if we can identify both[/latex] a_0  and  a_1[/latex] (which we can through the two stated initial conditions), then we can determine all of the remaining coefficients.

Specifically, since y(0) = 1 and a_{0} = y(0), it follows that a_{0} = 1. Similarly, with the given condition y′(0)=1 and the fact that a_{1} =y′(0), we know a_{1}=1.Thus, from the sequence of equalities with even indices above,

a_{0} =1, a_{2} =−\frac {1}{2}, a_{4} =− \frac {1}{4 · 3}a_{2} = \frac {1}{4 · 3 · 2}= \frac {1}{4!} ,
and a_{6} =− \frac {1}{30}a_{4} =− \frac {1}{6 · 5 · 4!}=− \frac {1}{6!}

From this and the stated recurrence relation for a_{2n} and a_{2n+2} , we observe that

a_{2n} = (−1)^{n} \frac {1}{(2n)!} , n = 0,1,2, . . . .             (8.3.5)

The formula (8.3.5) implies that the portion of the series expansion for y in which all of the powers of t are even will be

y_{1 }= 1− \frac {1}{2!} t^{2 }+ \frac {1}{4!} t^{4} − \frac {1}{6!} t^{6}+· · ·                 (8.3.6)

which we recognize as the familiar series expansion for cos t .

Returning to the recurrence relation involving the coefficients with odd indices, nearly identical work to that with the even coefficients shows that

a_{1} = 1, a_{3} =− \frac {1}{3!} , a_{5} =− \frac {1}{5 · 4}a_{3} = \frac {1}{5!} ,   and   a_{7} =− \frac {1}{42}a_{4} =− \frac {1}{7!}

These observations imply that the part of the expansion of y involving odd coefficients has form

y_{2} = t − \frac {1}{3!} t^{3} + \frac {1}{5!} t^{5} − \frac {1}{7!} t^{7}+· · · (8.3.7)

hich is sin t .

Hence our work with series expansions at (8.3.6) and (8.3.7) has shown that

y = 1+t − \frac {1}{2!} t^{2} − \frac {1}{3!} t^{3} + \frac {1}{4!} t^{4} + \frac {1}{5!} t^{5} − \frac {1}{6!} t^{6} − +· · ·
= 1−\frac {1}{2!} t^{2} + \frac {1}{4!} t^{4} − \frac {1}{6!} t^{6}+· · ·+t − \frac {1}{3!} t^{3} + \frac {1}{5!} t^{5} − \frac {1}{7!} t^{7}+· · ·
= cos t +sin t                  (8.3.8)