Lecture Notes in Differential Equations - Bruce E. Shapiro

256 LESSON 28. SERIES SOLUTIONS
If we can somehow find a non-trivial solution for the coefficients then we
have solved the initial value problem. In practice, we find the coefficients
by a generalization of the method of undetermined coefficients.
Example 28.1. Solve the separable equation
}
ty ′ − (t + 1)y = 0
y(0) = 0
(28.4)
by expanding the solution as a power series about the point t 0 = 0 and
show that you get the same solution as by the method of separation of
variables.
Separating variables, we can rewrite the differential equation as dy/y =
[(t + 1)/t]dt; integrating yields
ln |y| = t + ln |t| + C (28.5)
hence a solution is
y = cte t (28.6)
for any value of the constant c. We should obtain the same answer using
the method of power series.
To use the method of series, we begin by letting
y(t) =
∞∑
a k t k (28.7)
k=0
be our proposed solution, for some unknown (to be determined) numbers
a 0 , a 1 , . . . . Then
y ′ =
∞∑
ka k t k−1 (28.8)
k=0
Substituting (28.7) and (28.8) into (28.4),
∞∑
∞∑
0 = ty ′ − (t + 1)y = t ka k t k−1 − (t + 1) a k t k (28.9)
k=0
k=0