$v' = 1 - v^2$ ODE Solution

Question

My attempt at a solution involved separating both sides and eventually arriving at $v = tanh(t+c_1)$, which further evaluates to $v = \frac{e^{t+c}-e^{-t-c}}{e^{t+c}+e^{-t-c}}$, but WolframAlpha gives a different answer.

Answer 1

Your solution is correct $$v = \frac {e^{t+c}-e^{-t-c}}{e^{t+c}+e^{-t-c}}$$

Wolfram Alpha's solution is also correct. It is the constant and multiplication by $e^t$ which makes them look different.

Depending on how you represent your constant the solution changes its form but it is the same solution.

In WA's solution you have $e^2c$ which is just a constant but different from your constant. To avoid these kind of confusion it is best if you give your answer in terms of some initial condition.

Answer 2

you are missing some other solutions. The ones with $\tanh t$ and its translates are those where $-1 <v < 1$

Another solution is $v = -1$ Another still is $v = 1.$

The final ones look the same as far as formula. One type is $\frac{1}{\tanh t}$ for $t > 0$ only. There is an asymptote along the positive part of the $y$ axis. The ODE is autonomous, you may also translate.

The last is $\frac{1}{\tanh t}$ for $t < 0$ only. There is an asymptote along the negative part of the $y$ axis.