$y '(x) = a + \sqrt {y(x) - a \cdot x}$, to solve this differential equation, one can make a substitution: $ u = y(x)-ax $. For some arguments $u < 0$ could be true, and for others $u \geq 0$. Does that play a role when I want to determine $y(x)$?
Let $C_1,C_2,C$ be constants.
(i) If $u \geq 0$ for all $x$ then $\sqrt {y(x) - a \cdot x}=\sqrt {u}$, with the "seperation of variables"-method I get : $y(x)=(\frac{x}{2}+C)^2+ax$
(ii) If $u < 0$ for all $x$ then $\sqrt {y(x) - a \cdot x}=\sqrt{u} = \sqrt {(-1)\cdot (-u)}=i\sqrt {-u} $, with the "seperation of variables"-method I get :
$ \frac{du}{dx}=\frac{d(y-ax)}{dx}=\frac{dy}{dx}-a\leftrightarrow dy = du+ a\cdot dx \\ \rightarrow \frac{dy}{dx}=a+\sqrt{y-ax} \\ \leftrightarrow \frac{du+ a\cdot dx}{dx}=\frac{du}{dx}+a= a+\sqrt{u} \\ \leftrightarrow \frac{du}{dx}= \sqrt{u} = i\sqrt {-u} \\ \leftrightarrow i\cdot dx= \frac{du}{\sqrt{-u}}\\ i\cdot\int dx=i\cdot( x+C_1) = \int \frac{du}{\sqrt{-u}}=(-2)\sqrt{-u}+C_2 \\ \leftrightarrow \frac{ix}{2}+z = -\sqrt{-u}=- \sqrt{ax-y} \quad | \quad z:=\frac{iC_1-C_2}{2} \\ \leftrightarrow y= ax -(\frac{ix}{2}+z)^2$
If I put this "solution" in the differential equation, I can see that it's not right even though I don't know what I did wrong in this case?
(iii)If for some arguments $u < 0$ is true, and for others $u \geq 0$, then I don't know how to define a solution...
How do I know which $y(x)$ I need to calculate?
PS: I had very little contact with differential equations. If something isn't clear, please ask. English is not my native language, so sry for possible mistakes.