Page 288 - Schaum's Outline of Differential Equations

CHAP. 27] LINEAR DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS 271

Substituting these results into Eq. (27.5), with x replaced by t, we obtain the general solution

27.18. Find the general solution near x = 2ofy"-(x- 2)y' + 2y = 0.
To simplify the algebra, we first make the change of variables t = x — 2. From the chain rule we find the
corresponding transformations of the derivatives of y:

Substituting these results into the differential equation, we obtain

and this equation is to be solved near t = 0. From Problem 27.3, with x replaced by t, we see that the solution is

Substituting t = x — 2 into this last equation, we obtain the solution to the original problem as

27.19. Find the general solution near x = -1 of /' + xy' + (2x - l)y = 0.
To simplify the algebra, we make the substitution t = x —(—l)=x+1. Then, as in Problem, 27.18
2
2
(dyldx) = (dyldt) and ((fy/dx ) = (cPy/dt ). Substituting these results into the differential equation, we obtain

The power series solution to this equation is found in Problems 27.9 and 27.10 as