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The following two differential equations appear on page 93 of George Simmons' book on Differential Equations. While I have been able to solve them, I have some questions. [Not HW]

I. [tex](x^2-1)y'' - 2xy' + 2y = (x^2-1)^2[/tex]

II. [tex](x^2 + x)y'' + (2-x^2)y' - (2+x)y = x(x+1)^2[/tex]

How I solved them:

For the first equation, [itex]y = x[/itex] is a solution of the homogeneous part as can be verified by inspection. From this we can construct a second solution [itex]y = xv(x)[/itex] with [itex]v(x)[/itex] given by

[tex]v(x) = \int \frac{1}{x^2}e^{\int \frac{2x}{x^2-1}dx}dx = \left(x + \frac{1}{3x^3}\right)[/tex]

The homogeneous equation thus has the solution

[tex]y_{h}(x) = c_{1}x + c_{2}\left(x^2 + \frac{1}{3x^2}\right)[/tex]

For the general solution, add a polynomial and compare coefficients.

For the second equation, [itex]y = 1/x[/itex] is a solution of the homogenous part. The second solution is [itex]y = v(x)/x[/itex] where [itex]v(x)[/itex] is given by

[tex]v(x) = \int x^2 e^{-\int \frac{2-x^2}{x^2 + x} dx}dx = \int x^2 \left(\frac{1+x}{x^2}\right) e^{x}dx = \int (x+1)e^{x}dx = xe^{x}[/tex]

The general solution to the homogenous part is thus

[tex]y_{h}(x) = \frac{c_{1}}{x} + c_{2}e^{x}[/tex]

The general solution for the nonhomogeneous equation can be constructed as for the first part.

Now, I have two questions:

1. I could "guess" the simple solutions to the homogeneous equations in both cases, but how do I get to this solution rigorously? (I tried some substitutions but none of them worked.)

2. Is there some totally different way out to solve these equations too?

Thanks.

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# Alternate Method for solving two Second Order DEs.

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