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

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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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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.

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.

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.

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