Q:

Second order non linear differential equation: Central force question

The problem is as below:

I have derived that the particle satisfies the motion equation
$$ \frac{d^2u}{d \theta ^2 } + u = \frac{F(1/u)}{mh^2u^2} $$ 
by Newton's Law, $u= 1/r$ and $h = r^2 \frac{d \theta }{dt}$ is constant.  Hence, by substitution, we obtain from initial conditions that 
$h=av$, and that from the motion equation,
$$ \frac{d^2 u }{ d \theta ^2 } = 6au^2$$ 
And this as far I could get.  I could not really solve this equation - I tried multiplying by $u'$ both sides but nothing useful turns out.  Also I could not find a place to use the information on radial velocity. ( Probably used for solving the differential equation).  Any ideas?  
Thank you!   

A:

I know this is going to sound dumb, but all I knew how to do was to show that the candidate solution satisfied the initial conditions and the differential equation.  initially $r=a$, so $u=\frac1r=\frac1a$ as prescribed.  Then $\frac{dr}{dt}=-2v$, $\frac{d\theta}{dt}=\frac va$, so
$$\frac{dr}{d\theta}=-2a=-\frac1{u^2}\frac{du}{d\theta}=-r^2\frac{du}{d\theta}=-a^2\frac{du}{d\theta}$$
at $t=0$, and
$$\frac{du}{d\theta}=\frac d{d\theta}\frac1{a(1-\theta)^2}=\frac2{a(1-\theta)^3}$$
And this matches the initial condition $\frac{du}{d\theta}=\frac2a$. Then
$$\frac{d^2u}{d\theta^2}=\frac6{a(1-\theta)^4}=\frac{6a}{\left[a(1-\theta)^2\right]^2}6au^2$$
So no insights, but the solution just works.   
EDIT: Reviewing the problem, I can now see how close you were to solution.  You had
$$\frac{d^2u}{d\theta^2}\frac{du}{d\theta}=6au^2\frac{du}{d\theta}$$
Then you integrated to get
$$\frac12\left(\frac{du}{d\theta}\right)^2=2au^3+c_1$$
And here you were blocked because you didn't see how to apply the initial conditions to get $c_1=\frac12\left(\frac2a\right)^2-2a\left(\frac1a\right)^3=0$. From that point it's easy to wrap up
$$\frac{du}{u^{3/2}}=2a^{1/2}\,d\theta$$
$$-2u^{-1/2}=2a^{1/2}\theta+c_2$$
$$-2a^{1/2}=c_2$$
$$-2u^{-1/2}=-2r^{1/2}=2a^{1/2}(\theta-1)$$
$$r=a(1-\theta)^2$$

