find the solution of power series (2x^2-X^3)y''+(7x-6x^2)y'+(3-6x)y=0 near zero
Why do you call that a power series? It looks to me like a differential equation. Do you expect the solution to be a power series?
and I have a question also: is y a function of x, or something else?
To find the solution of the given power series near zero, we can use the method of Frobenius.
Step 1: Write the given differential equation in a standard form
(2x^2 - x^3)y'' + (7x - 6x^2)y' + (3 - 6x)y = 0
Step 2: Assume a power series solution of the form y = ∑(n=0 to ∞) a_n x^n
Substituting this into the differential equation, we get:
(2x^2 - x^3) ∑(n=0 to ∞) {n(n-1)a_n x^(n-2)} + (7x - 6x^2) ∑(n=0 to ∞) {na_n x^(n-1)} + (3 - 6x) ∑(n=0 to ∞) {a_n x^n} = 0
Step 3: Simplify the expression:
∑(n=0 to ∞) {2n(n-1)a_n x^n - n(n-1)a_n x^(n+1) - 7na_n x^n + 6na_n x^(n+1) + 6a_n x^n - 6a_n x^(n+1)} = 0
Step 4: Simplify the above equation by combining terms:
∑(n=0 to ∞) [(2n(n-1) - n(n-1))a_n - (7n - 6n + 6 - 6)a_n] x^n + ∑(n=0 to ∞) (6a_n - 6a_n)x^(n+1) = 0
Step 5: Collect like terms and separate the sums:
∑(n=0 to ∞) [(2n^2 - 2n - 7n + 6 + 6 - 6)a_n] x^n + x∑(n=0 to ∞) (6a_n - 6a_n)x^n+1 = 0
Step 6: Simplify further:
∑(n=0 to ∞) [(2n^2 - 9n + 6)a_n] x^n + x∑(n=0 to ∞) 0 = 0
Step 7: Simplify x∑(n=0 to ∞) 0 as 0.
Step 8: The equation now becomes:
∑(n=0 to ∞) (2n^2 - 9n + 6)a_n x^n = 0
Step 9: Set each term in the sum equal to zero to find a_n:
2n^2 - 9n + 6 = 0
Step 10: Solve the above quadratic equation to find n:
n = 1/2 or n = 3/2
Step 11: Using the values of n, find the corresponding values of a_n:
For n = 1/2, we have a_1/2 = 0
For n = 3/2, we have a_3/2 = 0
Step 12: The power series solution near zero is given by:
y = a_0 + a_1 x^(1/2) + a_2 x + a_3 x^(3/2) + ...
Therefore, the solution of the power series is y = a_0 + a_2 x + ...
Please note that this is a general solution at this stage. To obtain specific coefficients a_0, a_1/2, a_2, a_3/2, and so on, we would need additional initial/ boundary conditions.