Autonomous Differential Equation $y' = (1 - \frac{A}{y})^{-1}$ For an equation of the following form: $\frac{dy}{dt} = (1 - \frac{A}{y})^{-1}$ where $A > 0$ is a constant
Is there a way to solve for $y$ that doesn't involve using the Lambert $W$ function $($function such that $y = W(y)e^{W(y)})$?
When I go to solve this, I reach a stage with $\frac{y}{a} - \log(y) = \frac{t + c_1}{A}$ and I do not know of any ways to "isolate" the $y$ that don't involve the Lambert $W$ function.
 A: $$\frac{dy}{dt} = (1 - \frac{A}{y})^{-1}$$
$$(1 - \frac{A}{y})dy=dt$$
$$y-A\ln|y|=t+c$$
This is the solution on the form of implicit equation. Solving for $y$ cannot done with a finite number of elementary functions. It requires a special function namely the Lambert W function.
https://mathworld.wolfram.com/LambertW-Function.html
Details of the calculus :
$$e^y y^{-A}=e^{t+c}$$
$$y^{A}e^{-y}=e^{-(t+c)}$$
$$ye^{-\frac{y}{A}}=e^{-\frac{t+c}{A}}$$
$$-\frac{y}{A}e^{-\frac{y}{A}}=-\frac{1}{A}e^{-\frac{t+c}{A}}$$
Let $X=-\frac{y}{A}$ and $Y=-\frac{1}{A}e^{-\frac{t+c}{A}}$
$$Xe^X=Y\quad\to\quad X=W(Y)$$
$W(X)$ is the Lambert W function.
$$-\frac{y}{A}=W\left(-\frac{1}{A}e^{-\frac{t+c}{A}}\right)$$
$$y=-A\:W\left(-\frac{1}{A}e^{-\frac{t+c}{A}}\right)$$
A: No.
Otherwise you would have found a way to circumvent Lambert's function with a closed-form expression, and you can bet that this would have been discovered since the XVIIIth Century.
Functions where the variable appears both inside and outside of a transcendental function have the habit of not being invertible analytically. Note that Lambert's function has countably many branches (in the complex), and this is enough to show that it has no algebraic expression.
