1. Give the singular points for each of the following:
(a) (x - 3) y" + (x + 1)y = 0
(b) (x2 + 1) y"' + y" - x2 y = 0
Solutions:
Rewrite each in standard form:
(a) (x - 3) y" + (x + 1)y = 0
y" = - (x + 1) y / (x - 3)
The singular point is x = 3
(b) (x2 + 1) y"' + y" - x2 y = 0
y"' = - y" / (x + i) (x - i) + x2 y /(x + i) (x - i)
The singular points are x = + i
2. Determine whether x = 0 is an ordinary point of the differential equation:
x2 y " + 2 y' + xy = 0
Soln: Rewrite in standard form:
y " = - 2 y' / x2 - xy / x2
y " = - (2 / x2) y'- y / x
P(x) = 2 / x2
Q (x) = 1/x
Neither function is analytic at x= 0 so x = 0 is a singular point of the differential equation because the denominators go to 0.
3. Find the power series solution for the differential equation:
x y" + x3 y - 3 xy = 0
That satisfies: y = 0 and y' = 2 at x = 1
Soln.:
Rewrite in standard form:
y" = - x2 y + 3 y
Where: P(x) = x3 / x = x 2 and Q (x) = 3x/ x = 3
Here P(x) = x 2 and Q(x) = 3 are both polynomials and hence are analytic everywhere. Therefore every value of x is an ordinary point.
Differentiating the DE in successive steps:
y" = - x 2 y' + 3 y
y"' = - x 2 y" - 2x y' + y' + y
y iv = - x 2 y"' - 2 xy" - 2y' + y
Evaluate each derivative at x =1 :
y"(1) = -2 = - (1)2 (2) + 3(0)
y"'(1) = 0 = - (1)2 (-2) - 2(1)(2) + (2) - 0
y iv (1) = 0 = - (1)2 (0) - 2(1)(-2) - 2(2) + 0
Rem:
[(x - x 1 ) n D n + (x - x 1 ) b1 (x) D n-1 +
(x - x 1 ) n- 2 b1 (x) D n-2 +. .. .....(x - x 1 ) n- 1 b n- 1 (x) D + . . .
Then:
y(x) = 2(x - 1) - (x - 1) 2 + .. (x - 1) 1 + . . .
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