Show $\frac{du}{dx}=|u| \ \ (\ x\in\mathbb{R}$) has solution $u=Ae^x$

by Bell   Last Updated September 23, 2018 11:20 AM

Show the following ODE has solutions of the form $u=Ae^x$: $$\frac{du}{dx}=|u|, \ x\in\mathbb{R}.$$

My attempt:

I first considered the case where $u>0$. So, \begin{align} \frac{du}{dx}&=u \\ \int \frac{du}{u}&=\int dx \\ \ln|u|&=x+C, \ \ C\in\mathbb{R} \\ u&=Ae^x, \ \ A=e^c\in\mathbb{R} \end{align} Next, for the case where $u<0$ (using separations of variables as above), I get that, $$u=Ae^{-x}.$$ For the final case, $u=0$, $$u=0.$$ Only one of these cases admits the required form. Where have I gone wrong?

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