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

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• October 23rd 2010, 08:34 PM
weedingb
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
• October 23rd 2010, 08:43 PM
mr fantastic
Quote:

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
• October 23rd 2010, 09:16 PM
Jhevon
Quote:

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.
• October 24th 2010, 01:00 AM
chisigma
Quote:

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

$\chi$ $\sigma$