Can you saolve this please? thanks! #y''-2y'+y=e^x/sqrt(4-x^2)#

1 Answer
May 7, 2018

#y = e^(x)( x sin^(-1) (x/2) + sqrt(4-x^2) + alpha x + beta )#

Explanation:

#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 )#