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

Question

Cameron Alexander · Accepted Answer

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

#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 \ 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 )#

William Abdalla · Answer

undefined To solve the given differential equation $ y'' - 2y' + y = \frac{e^x}{\sqrt{4 - x^2}} $, we first solve the associated homogeneous equation $ y'' - 2y' + y = 0 $, which has characteristic equation $ r^2 - 2r + 1 = 0 $.

The characteristic equation has a repeated root $ r = 1 $, so the solution to the homogeneous equation is $ y_h(x) = (C_1 + C_2x)e^x $, where $ C_1 $ and $ C_2 $ are constants.

Now, we look for a particular solution to the non-homogeneous equation. We use the method of undetermined coefficients.

Given that the right-hand side is $ \frac{e^x}{\sqrt{4 - x^2}} $, we assume a particular solution of the form $ y_p(x) = Ae^x $, where $ A $ is a constant to be determined.

Taking derivatives and substituting into the original differential equation, we find $ A = \frac{1}{2} $.

Therefore, the particular solution is $ y_p(x) = \frac{1}{2}e^x $.

The general solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:

$$ y(x) = y_h(x) + y_p(x) = (C_1 + C_2x)e^x + \frac{1}{2}e^x $$

where $ C_1 $ and $ C_2 $ are constants determined by initial conditions, if provided.

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