Question 8.3.1: By assuming that y has a power series expansion of the form ...

Writing y(t ) = a_{0} +a_{1}t +a_{2}t^{2} +a_{3}t^{3}+· · ·, we know

y^{\prime}(t ) = a_{1} +2a_{2}t +3a_{3}t^{ 2} +4a_{4}t^{3}+· · ·

Equating y and y′, we observe that

a_{0} +a_{1}t +a_{2}t^{2} +a_{3}t^{3}+· · = a_{1} +2a_{2}t +3a_{3}t^{ 2} +4a_{4}t^{3}+· · ·                   (8.3.1)

Because of the uniqueness of Taylor series expansions (theorem 8.2.2), we may equate like coefficients of powers of t in (8.3.1), from which we deduce that the following recurrence relation among the coefficients ai must hold:

a_{0} = a_{1}
a_{1} = 2a_{2}
a_{2} = 3a_{3}

¦
a_{n} =(n+1)a_{n+1}

Provided that we know a_{0}, we can find all of the remaining values ofa_{i} . Clearly,a_{0}= y(0), so using the initial condition y(0) = 1,

a_{0} = 1, a_{1} = 1, a_{2} = \frac {1}{2}, a_{3} = \frac {1}{3}a_{2} = \frac {1}{3 · 2}, . . .

From this sequence of coefficients and the general recurrence relation a_{n+1} =\frac {1}{n+1}a_{n}, we observe that a_{n} = \frac {1}{n!} , and therefore

y(t ) = 1+t + \frac {1}{2!} t^{2} + \frac {1}{3!} t^{3}+· · ·+ \frac {1}{n!} t^{n} +· · ·

which we recognize as the familiar power series expansion of y(t ) = e^{t} , the solution to the IVP y′= y, y(0) = 1.