A simple looking yet tricky Second Order Nonlinear Differential Equation. I have been struggling with the following second order differential equation for quite a while,
$$y''(t)=\frac{(y'(t))^2}{y(t)}-\frac{(y'(0))^2}{y'(0)\times t+c}$$
Where $y'(0)$ is a constant of units $\frac{1}{[T]}$ and c is dimensionless. It looks fairly simple and I can bet there is a simple trick to be used to solve it but I just cannot see it. Any help would be greatly appreciated. If necessary, the initial conditions known are, $y(0)=1$ and $y'(0)$ is approximately $8.7e43$.
My first reflex was to write $y''(t)$ as $\frac{dy'(t)}{dt}$ since I could potentially just integrate $\int\frac{(y'(0))^2}{y'(0)\times t+c}dt$, which is easy. However it didn't work.
Simiralry, I have tried separating $y''(t)$ into $\frac{dy'(t)}{dy(t)}\times y'(t)$. This however didn't work since isolating $y'(t)dy'$ was out of my range of skills.
Finally, I just tried to integrate both sides with respects to $t$, however, I lack information to compute $\int\frac{(y'(t)^2)}{y(t)}dt$ when using integration by parts.
 A: For the dimensions in the denominator on the right hand side to work out, $c$ actually must be dimensionless. So I will proceed according to that assumption. Furthermore, this implies that $y$ is dimensionless.
Let $y'(0) = K$
BIG EDIT: I don't think my previous answer to the question holds up to scrutiny. In particular, the step where I mainly went wrong was a derivative integral swap that I don't think was valid. Therefore, this is my updated solution:
Let us define a system of equations by introducing a new function $v(t)$, independent of $y$.
Observe that if $v(t) = y'(t)$ AND $v' = \frac{vy'}{y}-\frac{K}{t+\frac{c}{K}}$, then the original equation is true. So we define this as our system of differential equations. To solve this system:
$$v'+\frac{K}{t+\frac{c}{K}} = \frac{v}{y}\frac{dy}{dt} $$
$$\left(v'+\frac{K}{t+\frac{c}{K}}\right)dt = \frac{v}{y}dy $$
$$\int\left(v'+\frac{K}{t+\frac{c}{K}}\right)dt = \int\frac{v}{y}dy $$
Because v is not a function of y,
$$v+K\ln(t+\frac{c}{K}) = v\ln(y)+M $$
We can pin down $M$ using the initial conditions of the problem: $y(0) = 1$, $v(0) = K$, so
$$K+K\ln(\frac{c}{K}) = M $$
Subbing back in:
$$v+K\ln(t+\frac{c}{K}) = v\ln(y)+K+K\ln(\frac{c}{K}) $$
$$K\left(\ln\left(t+\frac{c}{K}\right)-\left(1+\ln\left(\frac{c}{K}\right)\right)\right) = v(\ln(y)-1)$$
$$K\frac{\left(\ln\left(t+\frac{c}{K}\right)-\left(1+\ln\left(\frac{c}{K}\right)\right)\right)}{\ln(y)-1} = v$$
Because I don't want to deal with it and it's a constant, let's define the constant $A = \left(1+\ln\left(\frac{c}{K}\right)\right)$
$$K\frac{\left(\ln\left(t+\frac{c}{K}\right)-A\right)}{\ln(y)-1} = v$$
Substituting this into $y'(t) = v(t)$
$$K\frac{\left(\ln\left(t+\frac{c}{K}\right)-A\right)}{\ln(y)-1} = \frac{dy}{dt}$$
$$K\left(\ln\left(t+\frac{c}{K}\right)-A\right)dt = (\ln(y)-1)dy$$
$$\int K\left(\ln\left(t+\frac{c}{K}\right)-A\right)dt = \int (\ln(y)-1)dy$$
$$ K\left(\left(t+\frac{c}{K}\right)\ln\left(t+\frac{c}{K}\right)-\left((1+A)t+\frac{c}{K}\right)\right) = y(\ln(y)-2)+N$$
We know that $y(0)=1$, so to solve for N:
$$ K\left(\left(\frac{c}{K}\right)\ln\left(\frac{c}{K}\right)-\left(\frac{c}{K}\right)\right) = -2+N$$
$$ c\ln\left(\frac{c}{Ke}\right)+2 = N$$
So therefore,
$$ K\left(\left(t+\frac{c}{K}\right)\ln\left(t+\frac{c}{K}\right)-\left((1+A)t+\frac{c}{K}\right)\right)-c\ln\left(\frac{c}{Ke}\right)+2 = y(\ln(y)-2)$$
If you REALLY want to isolate $y$, you need to use the Lambert W function.
$$ \left(K\left(\left(t+\frac{c}{K}\right)\ln\left(t+\frac{c}{K}\right)-\left((1+A)t+\frac{c}{K}\right)\right)-c\ln\left(\frac{c}{Ke}\right)+2\right)\frac{1}{e^2} = \frac{y}{e^2}(\ln(\frac{y}{e^2}))$$
$$ W\left(\left(K\left(\left(t+\frac{c}{K}\right)\ln\left(t+\frac{c}{K}\right)-\left((1+A)t+\frac{c}{K}\right)\right)-c\ln\left(\frac{c}{Ke}\right)+2\right)\frac{1}{e^2}\right) = \frac{y}{e^2}$$
$$ e^2 W\left(\left(K\left(\left(t+\frac{c}{K}\right)\ln\left(t+\frac{c}{K}\right)-\left((1+A)t+\frac{c}{K}\right)\right)-c\ln\left(\frac{c}{Ke}\right)+2\right)\frac{1}{e^2}\right) = y$$
