- #1
VinnyCee
- 489
- 0
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!
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!
Last edited: