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

Answer 1

Cameron Alexander

#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 )#

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Answer 2

William Abdalla

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.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.