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Consider the differential equation $y' = 1 - y^2$.

First, is $y(x) = 1$ the only constant solution?

I now want to solve the equation for the initial value problem $y(0) = y_0$, with $y_0 > 1$.

Also, what's the maximal interval the solution function can be defined on? How does it behave at the edges (potentially for $x \to ± \infty$)?

Thanks in advance. Differential equations are new to me and I have trouble on how to visualize and solve an equation like this.

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closed as off-topic by Did, Daniel Robert-Nicoud, TravisJ, Daniel W. Farlow, graydad May 27 '15 at 2:31

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By setting $y'=0$ we easily get that the only constant solutions are given by $y=1$ and $y=-1$. This differential equation is separable and the general solution is given by: $$ y(x)=\frac{e^{2x}-K}{e^{2x}+K}$$ so, by plugging in $y(0)=y_0$, we get:

$$ y(x) = \frac{(1+y_0)\,e^{2x}-(1-y_0)}{(1+y_0)\,e^{2x}+(1-y_0)}.$$

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  1. No: try $-1$.
  2. Hint: separable differential equation.
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