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
I got your answer, and it is equivalent to the answer shown in the link provided by Mr. F. What was the answer you got from someone else? Ah, it doesn't matter, your answer's fine.

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 $y^{'} = 1-y^{2}$ is of the type 'separable variables' and its solution seems at first very confortable...

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

A subtil pitfall however exists because is...

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

... so that the solution is...

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

Kind regards

$\chi$ $\sigma$