I would like to solve:
$ \frac{d^2y}{dx^2} - \frac{2}{y^2}=0$
with $y(0)=a$ and $y'(0)=0$
Where $a$ is a known constant.
Thanks in advance.
I would like to solve:
$ \frac{d^2y}{dx^2} - \frac{2}{y^2}=0$
with $y(0)=a$ and $y'(0)=0$
Where $a$ is a known constant.
Thanks in advance.
Change variables so that the independent variable is $y$ and the dependent on $p=y'$. Then $p\dot p=y''$ (dots stand for derivatives w.r.t. $y$, primes w.r.t. $x$) and the equation becomes $p\dot p-\frac2{y^2}=0$ or $\frac{d}{dy}(\frac{p^2}2)=\frac2{y^2}.$ This can be solved by integrating into a first order equation for $y$ as a function of $x$.
Maybe this is the same technique, but what I have done before for y''=f(y) is treated as \frac{{\rm d}y'}{{\rm d}x} = f(y) \frac{{\rm d}y'}{{\rm d}y} \frac{{\rm d}y}{{\rm d}x} = f(y) y'\,\frac{{\rm d}y'}{{\rm d}y} = f(y) \int y'\,{\rm d}y' =\int f(y)\,{\rm d}y +K_0 \frac{1}{2} y'^2 = g(y) with $g(y)=\int f(y)\,{\rm d} y + K_0$ y' = \sqrt{2\,g(y)} $ \frac{{\rm d}y}{\sqrt{2\,g(y)}} = {\rm d}x$ $ x = \int \frac{1}{\sqrt{2\,g(y)}}\,{\rm d}y+K_1 $
With your example $f(y)=2/y^2$ $g(y)=\int 2/y^2\,{\rm d}y+K_0 = -2/y +K_0 $ $ x = \int \frac{1}{\sqrt{2\,\left(-2/y +K_0\right)}}\,{\rm d}y+K_1 $ and with the IC
given $ x = \frac{\sqrt{a}}{4} \left( 2 a \ln{\left(\sqrt{y-a}+\sqrt{y}\right)}-a \ln{a}+2 \sqrt{y (y-a)} \right) $
Hint $\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$ Multiply by $2y'$ $2y'\frac{d^2y}{dx^2} - \frac{4y'}{y^2}=0$ $((y')^2)' +4 \left(\frac{1}{y}\right)'=0$ Then integrate $(y')^2 +\frac{4}{y}=K_1$ $.......$