I need Help with Solving: $\int{y(t)*\ddot{y(t)}}dt$? As I was solving a differential equation I came across this integral which I have no idea how to solve, I managed to reduce the equation to : 
$$Ct = y(t)\dot{y}(t)-y(t)$$
where $C$ is just a constant. Having never taken a course in differential equations, I don't know how to solve this if you could please help, thank you.
The original differential equation is: 
$$\frac{C}{y(t)} =\ddot{y(t)}$$ 
where $C\ne 0$ and $y(0)=R > 0$
I have tried the method of writing 
$$\ddot{y}(t)= \frac{d\dot{y}(t)}{dt}$$
which is just 
$$\frac{d\dot{y}(t)}{dt}\frac{dy(t)}{dy(t)}$$
which is just 
$$\dot{y}(t)\frac{d\dot{y}(t)}{dy(t)}$$
which gives the equation
$$\frac{C}{y(t)}=\frac{d\dot{y}(t)}{dy(t)}\dot{y}(t)$$
then to multiply both sides by $dy(t)y(t)$ 
then integrate to get the question: 
$$t = \int_{R}^0{\frac{1}{\sqrt{2C\ln(y)}}}dy$$ 
but this integral has an imaginary solution. This problem arose during a physics equation I was doing where I calculated the time until an object who is affected by the magnetic field of a wire collides with the wire, and so the solution must be real. A solution is always welcome, thank you.
 A: Considering the original equation
$$y''=\frac C {y}$$ you need to use first the fact that
$$\frac{d^2t}{dy^2}=-\frac{\dfrac{d^2y}{dt^2}}{\left(\dfrac{dy}{dt}\right)^3}\implies \dfrac{d^2y}{dt^2}=-\frac{\dfrac{d^2t}{dy^2}}{\left(\dfrac{dt}{dy}\right)^3}$$ making the equation to be
$$-\frac{t''}{(t')^3}=\frac C y$$
Now, reduce the order using $p=t'$ to get
$$\frac {p'}{p^3}=-\frac C y$$ which is easy to integrate
$$p=\pm \frac{1}{\sqrt{2 C \log (y)+ c_1}}$$ where $c_1$ has to be fixed by some initial condition.
Now, comes the tedious part : integrating again
$$t+c_2=\pm\sqrt{\frac{\pi }{2C}}\, e^{-\frac{c_1}{2 C}}\, \text{erfi}\left(\frac{\sqrt{2 C \log (y)+c_1}}{\sqrt{2 C} }\right)$$
A: Let $u = \dot y$. Then $\ddot y = \dfrac{du}{dt} = \dfrac{du}{dy}\dfrac{dy}{dt} = u \dfrac{du}{dy}$
So you have
$$ u \frac{du}{dy} = \frac{C}{y} $$
Integrating both sides gives
$$ \frac{u^2}{2} = C\ln y + A $$
where $A$ is the integration constant
You're missing a second initial condition. The value of $\dot y(0)$ would help us find $A$.
After that, you can find the inverse function using separation of variables
$$ t = \pm\int_R^y \frac{ds}{\sqrt{2(C\ln s + A)}} ds $$
Again, the sign of $\dot y(0)$ would determine the sign of the square root
A: In the case of differential equation $y(t)y''(t)=\alpha t$ you have asked in your recent question Anyone know how to solve this differential equation $yy''=\alpha t$, I am afraid that there are no closed form formula.
Nevertheless, I have obtained interesting results by simulation with one of the Runge-Kutta blackboxes available on Matlab. Here are the curves of $y=f(t)$ for different values of $f(0)$, all with $f'(0)=0$ (Matlab program below).


function diffequ;
ts = 0:0.01:20; % time sampling 
sed = @(t,X) SED(t,X); % call to SED
for k=-10:10
    X0 = [10*k;0]; 
    [t,X] = ode45(sed, ts, X0); % a version of Runge Kutta
    plot(t,X(:,1));hold on;
end;
%
function XP = SED(t,X);% 
p=X(1);q=X(2); % p=f, q=f'
pp=q; % q:=p'
qp=10*t./p; % p''=q'=a*t/p;
XP=[pp;qp];


