- #1
jackmell
- 1,807
- 54
Hi,
I derived the equation:
[tex]1+(y')^2-y y''-2y\left(1+(y')^2\right)^{3/2}=0[/tex]
Letting [itex]y'=p[/itex] and [itex]y''=p\frac{dp}{dy}[/itex], I obtain:
[tex]\frac{dp}{dy}=\frac{1+p^2-2y(1+p^2)^{3/2}}{yp}[/tex]
I believe it's tractable in p because Mathematica gives a relatively simple answer:
[tex]p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}\\
\\
-\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}
\end{cases}[/tex]
I don't see how to get to that answer and I was wondering if someone here could help me with this.
Thanks,
Jack
I derived the equation:
[tex]1+(y')^2-y y''-2y\left(1+(y')^2\right)^{3/2}=0[/tex]
Letting [itex]y'=p[/itex] and [itex]y''=p\frac{dp}{dy}[/itex], I obtain:
[tex]\frac{dp}{dy}=\frac{1+p^2-2y(1+p^2)^{3/2}}{yp}[/tex]
I believe it's tractable in p because Mathematica gives a relatively simple answer:
[tex]p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}\\
\\
-\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}
\end{cases}[/tex]
I don't see how to get to that answer and I was wondering if someone here could help me with this.
Thanks,
Jack