#y''-2y'+y=e^x/sqrt(4-x^2)#
This is:
#(D-1)^2 y =e^x/sqrt(4-x^2)#, with operator #D = d/(dx)#
Let #z = (D-1) y#:
#(D-1) z =e^x/sqrt(4-x^2)#
Or:
#z'- z =e^x/sqrt(4-x^2)#
Integrating factor: #mu(x) = e^(- int\ dx ) = Ae^(-x)#
#mu( z'- z =e^x/sqrt(4-x^2))#
#implies (z e^(-x))^' =1/sqrt(4-x^2)#
#z e^(-x) = int \ dx \ 1/sqrt(4-x^2) qquad square#
With #x = 2 sin A#, the RHS is:
- #int \ d(2 sin A) \ 1/sqrt(4-4 sin^2 A)#
# = int \ dA \ (2 cos A )/(2cos A) = A + " const" = sin^(-1) (x/2) + " const" #
#square# becomes:
#z = (D-1) y = e^x sin^(-1) (x/2) + alpha e^x #
This is:
#y' - y = e^x sin^(-1) (x/2) + alpha e^x#
With the same Integrating factor: #mu(x) = e^(- int\ dx ) = Ae^(-x)#
#mu(y' - y = e^x sin^(-1) (x/2) + alpha e^x)#
#implies y e^(-x) = int dx qquad sin^(-1) (x/2) + alpha qquad circ #
Second part is trivial of RHS, and first part can be IBP'd using the result that follows #square#.
# int dx \ (x)^' sin^(-1) (x/2)#
# x sin^(-1) (x/2) - int dx \ x (sin^(-1) (x/2))^'#
From the preceding integration:
# = x sin^(-1) (x/2) - int dx \ x/sqrt(4-x^2)#
Pattern matching:
# = x sin^(-1) (x/2) - int dx \ (-sqrt(4-x^2) )^'#
# = x sin^(-1) (x/2) + sqrt(4-x^2) + beta #
#circ# now reads:
#y e^(-x) =x sin^(-1) (x/2) + sqrt(4-x^2) + alpha x + beta#
#implies y = e^(x)( x sin^(-1) (x/2) + sqrt(4-x^2) + alpha x + beta )#
