Using the Lambert W to express a solution of a differential equation.

Question

I solved a differential equation some time ago and I need to solve for $y$. How can we solve for $y$ using the Lambert W function?

$$C_1+x = e^y+Cy$$

Answer 1

Hint

$$\large C^{-1}\exp\left(C^{-1}e^y+y\right)=(C^{-1} e^y)e^{(C^{-1} e^y)} $$

Answer

$$\hskip 2in \displaystyle \begin{array}{} A+x=e^y+Cy \\ \frac{A+x}{C}=\frac{e^y}{C}+y \\ \frac{1}{C}\exp\left(\frac{A+x}{C}\right)=\frac{e^y}{C}\exp\left(\frac{e^y}{C}\right) \\ W\left(\frac{1}{C}e^{(A+x)/C}\right)=\frac{e^y}{C} \\ y=\log\left[ C\; W\left(\frac{1}{C}e^{(A+x)/C}\right)\right] \end{array}$$

Note also that $\log W(z)=\log z-W(z)$, if you want to compare with what W|A gives.