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


which is just


which gives the equation


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.


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 $$t+c_2=\pm\sqrt{\frac{\pi }{2C}}\, e^{-\frac{c_1}{2 C}}\, \text{erfi}\left(\frac{\sqrt{2 C \log (y)+c_1}}{\sqrt{2 C} }\right)$$


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


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).

enter image description here

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;
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;

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