I solved a simple separable (and also linear) first-order ODE as follows: $$ \begin{align} \frac{dy}{dx}&=x(1-y) \\ \frac{1}{1-y}\frac{dy}{dx}&=x \\ \int\frac{1}{1-y}\frac{dy}{dx}dx&=\int xdx \\ \ln|1-y|&=\frac{x^2}{2}+C\\ y&=1-e^{\frac{x^2}{2}+C}. \end{align} $$
However, online calculators suggest that the exponent in the solution is negative: $$ y=1-e^{-\frac{x^2}{2}-C}. $$
I'm sure the mistake is trivial, but I can't see it! Where am I going wrong?
$\int \frac{1}{1-y}dy=\int x dx$
put $u=1-y$
$\int -u^{-1} du=\int x dx$
$-\ln (u)=\frac{x^2}{2}+C$
$u=e^{-\frac{x^2}{2}+ C}$
$1-y=e^{-\frac{x^2}{2}+ C}$
$y=1-e^{-\frac{x^2}{2}+ C}$