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I am puzzled over the following differential equation: $$y' ^2+(y-1)y'-y=0$$ I tried solving for $y$ and got $$y=\frac{y'(y'-1)}{1-y'}$$ But I am not sure where that takes me or if that was even the right approach. Grateful for any tips!

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2 Answers 2

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The given equation can be written as $(y'+y)(y'-1)=0$, so either $y'=-y$ or $y'=1$, so $y=Ce^{-x}$ and $y=x+C$ are solutions.

Another solution is a piecewise combination of the above. At any point where we switch between the two forms, we have $-y = y' = 1$, so $Ce^{-x}=-1$. This can only occur if $C<0$, at $x=\log(-C)=c$, giving the following additional solutions:

$$y(x) = \begin{cases} -e^{c-x} & x \le c \\ x-c-1 & x \ge c \end{cases}$$ and $$y(x) = \begin{cases} x-c-1 & x \le c \\ -e^{c-x} & x \ge c \end{cases}$$

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    $\begingroup$ Thank you. Got it now! $\endgroup$
    – Iris
    Commented Jul 3, 2019 at 0:18
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$$y=\frac{y'(y'-1)}{-(y'-1)}=-y'$$ Thus $y=c e^{-x}$ for some $c\in \mathbb R$.

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    $\begingroup$ This misses the $y' = 1$ solution $\endgroup$
    – GFauxPas
    Commented Jul 3, 2019 at 0:08

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