Find y as a function of x if x2 y 19xy 81y 0 y1 9 y 1 2

Find y as a function of x if x^2 y\'\' + 19xy\' + 81y = 0, y(1) = 9, y\' (1) = -2. y = _________

Solution

Solve ( d^2 y(x))/( dx^2) x^2+19 ( dy(x))/( dx) x+81 y(x) = 0:
Assume a solution to this Euler-Cauchy equation will be proportional to x^ for some constant .
Substitute y(x) = x^ into the differential equation:
x^2 ( d^2 )/( dx^2)(x^)+19 x ( d)/( dx)(x^)+81 x^ = 0
Substitute ( d^2 )/( dx^2)(x^) = (-1) x^(-2) and ( d)/( dx)(x^) = x^(-1):
^2 x^+18 x^+81 x^ = 0
Factor out x^:
(^2+18 +81) x^ = 0
Assuming x!=0, the zeros must come from the polynomial:
^2+18 +81 = 0
Factor:
(+9)^2 = 0
Solve for :
= -9 or = -9
The multiplicity of the root = -9 is 2 which gives y1(x) = c1/x^9, y2(x) = (c2 log(x))/x^9 as solutions, where c1 and c2 are arbitrary constants.
The general solution is the sum of the above solutions:
Answer: |
| y(x) = y1(x)+y2(x) = c1/x^9+(c2 log(x))/x^9

this is the general solution

in case we need the particular solution we can substitute the value of y(1) and y\'(1).