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__Problem 1__:

Find solution for:

[tex](x^2\,-\,4)\,y''\,+\,(3\,x)\,y'\,+\,y\,=\,0[/tex]

using power series methods.

__Answer 1__:

I get a recursion formula:

[tex]a_{n\,+\,2}\,=\,\frac{n\,+\,1}{4\,(n\,+\,2)}\,a_n[/tex]

and a final answer:

[tex]y(x)\,=\,a_0\,\left[1\,+\,\frac{x^2}{8}\,+\,\frac{3}{128}\,x^4\,+\,\frac{5}{1024}\,x^6\,+\,...\right]\,+\,a_1\,\left[x\,+\,\frac{x^3}{6}\,+\,\frac{x^5}{30}\,+\,\frac{x^7}{140}\,+\,...\right][/tex]

Does that look right?

__Problem 2__:

Use Euler's method to solve:

[tex](2\,x^2)\,y''\,+\,(x)\,y'\,+\,y\,=\,0[/tex]

__Answer 2__:

Using the quadratic equation to solve for r:

[tex]2\,r^2\,-\,r\,+\,1\,=\,0[/tex]

[tex]r\,=\,\frac{1}{4}\,\pm\,\frac{\sqrt{7}}{4}\,i[/tex]

Which means that:

[tex]\lambda\,=\,\frac{1}{4}[/tex] AND [tex]\mu\,=\,\frac{\sqrt{7}}{4}[/tex]

And finally:

[tex]y(x)\,=\,C_1\,x^{\frac{1}{4}}\,cos\,(\frac{\sqrt{7}}{4}\,ln\,x)\,+\,C_2\,x^{\frac{1}{4}}\,sin\,(\frac{\sqrt{7}}{4}\,ln\,x)[/tex]

Thanks for the checking in advance!

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