Why Lambert answer doesn't satisfy the original equation in WolframAlpha

Question

I solved the differential equation $y(y'+a)=b$ and found the answer $$ay+b\ln(y-\frac{b}{a})=-a^2(t+c)$$ as I wanted the explicity form of my solution, by giving it to WolframAlpha at address https://www.wolframalpha.com/input?i=y%28y%27%2Ba%29%3Db it is in terms of Lambert function: $$y(x) = \frac{b}{a} \left(W\left(-\frac{1}{b}e^{-\frac{xa^2}{b} + \frac{Ca^2}{b} - 1}\right) + 1\right)$$ but my handy answer is \begin{align} u=y-\frac{b}{a} \hspace{1cm} \Rightarrow \hspace{1cm} & b\ln u=-a^2t-a^2C-au-b\\ & u=e^{-\frac{a^2t}{b} - \frac{a^2C}{b} - 1}e^{-\frac{a}{b}u}\\ & \frac{a}{b}u\ e^{\frac{a}{b}u}=\frac{a}{b}\ e^{-\frac{a^2t}{b} - \frac{a^2C}{b} - 1}\\ & \frac{a}{b}u=W\left(\frac{a}{b}\ e^{-\frac{a^2t}{b} - \frac{a^2C}{b} - 1}\right)\\ y=u+\frac{b}{a} \hspace{1cm} \Rightarrow \hspace{1cm} & y(x) = \frac{b}{a} \left(W\left(\frac{\color{red}{a}}{b}e^{-\frac{xa^2}{b} + \frac{Ca^2}{b} - 1}\right) + 1\right) \end{align} which is differ from Wolfram answer. Where is my mistake?