(x^2+1) y^''+2xy^'=0
Solve for two power series solution about ordinary point x=0
why the stupid carets? These are not exponents.
(x^2+1)y"+2xy'=0
Since
y = c1 + c2 arctan(x)
I suspect you can now come up with the series.
To solve the differential equation (x^2+1) y^'' + 2xy^' = 0 using power series, we can express the solution y(x) as a power series about the ordinary point x = 0. Let's proceed step by step:
Step 1: Assume that y(x) can be expressed as a power series:
y(x) = ∑[n=0 to ∞] ( a[n] * x^n ),
where a[n] are constants to be determined and x is the independent variable.
Step 2: Determine the derivatives y'(x) and y''(x):
y'(x) = ∑[n=0 to ∞] ( a[n] * n * x^(n-1) ) = ∑[n=1 to ∞] ( a[n] * n * x^(n-1) )
= ∑[n=0 to ∞] ( a[n+1] * (n+1) * x^n ),
y''(x) = ∑[n=0 to ∞] ( a[n+1] * (n+1) * (n+1) * x^n ).
Step 3: Substituting y(x), y'(x), and y''(x) into the differential equation, we have:
(x^2 + 1) * ∑[n=0 to ∞] ( a[n+1] * (n+1) * (n+1) * x^n ) + 2x * ∑[n=0 to ∞] ( a[n+1] * (n+1) * x^n ) = 0.
Step 4: Simplify the expression and collect like terms:
∑[n=0 to ∞] ( a[n+1] * (n+1) * (n+1) * x^(n+2) ) + ∑[n=0 to ∞] ( a[n+1] * (n+1) * x^(n+1) )
+ 2x * ∑[n=0 to ∞] ( a[n+1] * (n+1) * x^n ) = 0.
∑[n=2 to ∞] ( a[n-1] * (n-1) * (n-1) * x^n ) + ∑[n=1 to ∞] ( a[n] * n * x^n )
+ 2x * ∑[n=0 to ∞] ( a[n+1] * (n+1) * x^n ) = 0.
Step 5: Equate the coefficients of each power of x to zero:
For n ≥ 2: a[n-1] * (n-1) * (n-1) = 0. This gives a[n-1] = 0 for n ≥ 2.
For n ≥ 1: a[n] * n + 2a[n+1] * (n+1) = 0. Simplifying, we get a[n+1] = - (a[n] * n) / (2 * (n+1)).
Step 6: Determine the values of a[0] and a[1]:
From the above recurrence relation, we have a[1] = - (a[0] * 1) / (2 * (1+1)) = - a[0] / 4.
Step 7: Write out the power series solution:
y(x) = a[0] + a[1] * x + a[2] * x^2 + ...
Substituting the values from Step 6, we have:
y(x) = a[0] - a[0] * x/4 + a[2] * x^2 + ...
Simplifying, we get:
y(x) = a[0] * (1 - x/4) + a[2] * x^2 + ...
Thus, we have derived the power series solution to the given differential equation about the ordinary point x = 0.