Answer:
(a) To find the general solution of the differential equation t^2y'' + ty' - 9y = 0:
Step 1: Assume a power series solution of the form y = ∑(n=0 to ∞) a_nt^n.
Step 2: Calculate the first and second derivatives of y:
y' = ∑(n=0 to ∞) na_nt^(n-1)
y'' = ∑(n=0 to ∞) n(n-1)a_nt^(n-2)
Step 3: Substitute the power series solution and its derivatives into the differential equation:
∑(n=0 to ∞) n(n-1)a_nt^(n-2) + t∑(n=0 to ∞) na_nt^(n-1) - 9∑(n=0 to ∞) a_nt^n = 0
Step 4: Simplify the equation and combine like terms:
∑(n=0 to ∞) [n(n-1)a_nt^(n-2) + tna_nt^(n-1) - 9a_nt^n] = 0
Step 5: Adjust the indices of the summation:
∑(n=2 to ∞) [(n(n-1)a_n + tna_n - 9a_n)t^(n-2)] + a_0 + a_1t = 0
Step 6: Equate the coefficients of like powers of t to zero:
a_0 + a_1t = 0 (for n = 0, 1)
(n(n-1)a_n + tna_n - 9a_n)t^(n-2) = 0 (for n ≥ 2)
From the equation a_0 + a_1t = 0, we can solve for a_0 and a_1:
a_0 = 0
a_1 = 0
For n ≥ 2, the equation (n(n-1)a_n + tna_n - 9a_n)t^(n-2) = 0 gives:
n(n-1)a_n + tna_n - 9a_n = 0
Simplifying the equation, we get:
a_n[n(n-1) + tn - 9] = 0
Since this equation holds for all values of t, we have two cases:
Case 1: a_n = 0 (for n ≥ 2)
This implies that all terms in the power series beyond the first two terms are zero.
Case 2: n(n-1) + tn - 9 = 0
This is a quadratic equation in terms of n. Solving it will give us the specific values of n.
The general solution of the differential equation is the combination of the power series solutions from both cases. However, since the specific values of n are not provided, we cannot provide the exact form of the general solution without solving the quadrati
Step-by-step explanation: