determine a series if solution of the equation y"(x)+x.y'(x)+y(x)=0
show me step plz
review the conditions for the series to exist. In this case, it will exist for x=0, so we want
y = ∑anx^n
y' = ∑nanx^(n-1)
y" = ∑n(n-1)anx^(n-2)
The DE then says
∑n(n-1)anx^(n-2) + x*∑nanx^(n-1) + ∑anx^n = 0
∑n(n-1)anx^(n-2) + ∑nanx^n + ∑anx^n = 0
∑n(n-1)anx^(n-2) + ∑(n+1)anx^n = 0
Now shift y" so the powers of x add up
∑(n+2)(n+1)an+2x^n + ∑(n+1)anx^n = 0
∑[(n+2)(n+1)an+2 + (n+1)an]x^n = 0
For the series to be zero for all x, we have the recurrence relation
(n+2)(n+1)an+2 + (n+1)an = 0
an+2 = -an/(n+2)
At this point, just start listing the coefficients and see where you go. There is a similar DE used as an example at
http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx
To find the solution of the differential equation y''(x) + x*y'(x) + y(x) = 0, we can use the power series method.
Step 1: Assume a power series solution of the form y(x) = Σa_n*x^n, where a_n are coefficients to be determined.
Step 2: Differentiate y(x) twice to find y'(x) and y''(x).
y(x) = a_0 + a_1*x + a_2*x^2 + ...
y'(x) = a_1 + 2*a_2*x + 3*a_3*x^2 + ...
y''(x) = 2*a_2 + 6*a_3*x + 12*a_4*x^2 + ...
Step 3: Substitute the power series solutions into the differential equation and collect terms with the same powers of x.
y''(x) + x*y'(x) + y(x) = (2*a_2 + 6*a_3*x + 12*a_4*x^2 + ...)
+ x*(a_1 + 2*a_2*x + 3*a_3*x^2 + ...)
+ (a_0 + a_1*x + a_2*x^2 + ...) = 0
Step 4: Equate the coefficients of the powers of x to zero to obtain a set of recursive equations.
For the coefficient of x^0:
2*a_2 + a_0 = 0 --> equation (1)
For the coefficient of x^1:
2*a_2 + a_1 = 0 --> equation (2)
For the coefficient of x^2:
6*a_3 + 2*a_2 + a_0 = 0 --> equation (3)
For the coefficient of x^3:
6*a_3 + 2*a_2 + a_1 = 0 --> equation (4)
And so on, for higher powers of x.
Step 5: Solve the set of recursive equations (1)-(4) to find the coefficients a_n.
Starting with equation (1):
a_2 = -a_0/2
Using a_2 in equation (2):
a_1 = -2*a_2 = a_0
Using a_1 in equation (4):
6*a_3 + 2*a_2 + a_1 = 0
6*a_3 + 2*(-a_0/2) + a_0 = 0
6*a_3 = 0
a_3 = 0
Continuing in this manner, we see that all the odd coefficients are zero.
For the even coefficients, using the general equation (n > 1):
a_n = (-1/n) * (a_(n-2))
For example, for a_4:
a_4 = -1/4 * (a_2) = -1/4 * (-a_0/2) = a_0/8
So, the general solution for y(x) is:
y(x) = a_0 * (1 + (x^2)/8 + (x^4)/384 + ...)
This is the power series solution for the given differential equation.
Note that a_0 is an arbitrary constant, and it can be selected freely.