What is the analytic solution for $y' = -\frac{a}{y} + \frac{b}{y^2}$?

Question

For $y: \mathbb{R}_+ \to \mathbb{R}_+$, if $y'(x) = -\frac{a}{y} + \frac{b}{y^2}$, then $y(x) = ?$

Can anyone give an exact solution of this type?

Update: Yes, $a,b$ are some positive constants. By separating the variables and integrating over $x,y$ we have \begin{align} \frac{y^2}{-ay + b} dy &= dx \\ -\frac{2 b^2 \log(b-ay) + a^2 y^2 + 2 a b y}{2 a^3} &= x \end{align}

For this can we get the explicit formula for $y(x)$?

Answer 1

Hint: Since $-\frac ay+\frac b{y^2}$ is a function of $y$, your differentiable equation is separable.

Answer 2

Suppose $y'=-\frac ay+\frac b{y^2}$. Then,

$$dx=\frac{y^2}{b-ay}\,dy$$

Now integrate and obtain an implicit equation for $y$.