# Thread: y' = 1 - y^2 Nonlinear 1st order ODE help

1. ## y' = 1 - y^2 Nonlinear 1st order ODE help

Hi,
I'm trying to solve y' = 1 - y^2.

Using partial fractions I ended up with y = (Ae^2x - 1)/(Ae^2x +1). However, this differs to an answer I was given by someone else. Am I correct?

Cheers,
Ben

2. Originally Posted by weedingb
Hi,
I'm trying to solve y' = 1 - y^2.

Using partial fractions I ended up with y = (Ae^2x - 1)/(Ae^2x +1). However, this differs to an answer I was given by someone else. Am I correct?

Cheers,
Ben
solve dy&#47;dx &#61; 1 - y&#94;2 - Wolfram|Alpha

3. Originally Posted by weedingb
Hi,
I'm trying to solve y' = 1 - y^2.

Using partial fractions I ended up with y = (Ae^2x - 1)/(Ae^2x +1). However, this differs to an answer I was given by someone else. Am I correct?

Cheers,
Ben

4. Originally Posted by weedingb
Hi,
I'm trying to solve y' = 1 - y^2.

Using partial fractions I ended up with y = (Ae^2x - 1)/(Ae^2x +1). However, this differs to an answer I was given by someone else. Am I correct?

Cheers,
Ben
The first order DE $\displaystyle y^{'} = 1-y^{2}$ is of the type 'separable variables' and its solution seems at first very confortable...

$\displaystyle \displaystyle \frac{dy}{1-y^{2}} = dx$ (1)

A subtil pitfall however exists because is...

$\displaystyle \displaystyle \int \frac{dy}{1-y^{2}}=\left\{\begin{array}{ll}\tanh^{-1} y ,\,\,|y|< 1\\{}\\ \coth^{-1} y ,\,\, |y|>1\end{array}\right.$ (2)

... so that the solution is...

$\displaystyle \displaystyle y(x) =\left\{\begin{array}{ll}\tanh (x+c) ,\,\,|y(0)|< 1\\{}\\ \coth (x+c) ,\,\, |y(0)|>1\end{array}\right.$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$