# Differential Equation (help with the algebra)

1. Apr 7, 2012

### Nano-Passion

1. The problem statement, all variables and given/known data[/b]
I need help knowing how my textbook jumped from one equation to the other in the example given.
$$\frac{y-2}{y+2}= e^{4x+c}$$
They solved for y and got
$$y = 2\frac{1+ce^4x}{1-ce^4x}$$

What is the intermediate step? I tried but couldn't isolate y.

And then they do something again that I don't know how they got:

"Now if we factor out the right hand sign of the differential equation as $$dy/dx = (y-2)(y+2)$$

I couldn't reach that step either.

2. Apr 7, 2012

### Office_Shredder

Staff Emeritus
$$\frac{y-2}{y+2}=A$$
becomes
$$y-2 = (y+2)A = Ay+2A$$

putting all the y's together gives
$$y-Ay=2+2A$$
factor out a y
$$y(1-A)=2+2A$$
and solve
$$y=\frac{2+2A}{1-A}$$

Note that this is NOT what you tex'd up, you have a typo on either your first or second equation

For the differential equation I think you're going to have to tell us what it looked like before they wrote it like that for us to tell you what they did

3. Apr 7, 2012

### Nano-Passion

I've checked and rechecked meticulously and I can assure you that there were no mistakes. The only thing that I omitted was the ± sign for e.

$$\frac{y-2}{y+2}= ±e^{4x+c}$$

4. Apr 7, 2012

### Nano-Passion

Here is the whole problem:

$$\frac{dy}{y^2-4}=dx$$
Using partial fractions
$$\frac{1/4}{y-2}-\frac{1/4}{y+2}dy=dx$$
Integrating both sides
$$1/4 (ln(y-2) - ln (y+2)) = x$$
Logarithmic property gives
$$\frac{1}{4} \frac{ln(y-2)}{y+2} = x + c$$
Raising everything to e we get
$$\frac{y-2}{y+2}= ±e^{4x+c}$$
Solving for y
$$y = 2\frac{1+ce^4x}{1-ce^4x}$$
"Now if we factor out the right hand sign of the differential equation as $$dy/dx = (y-2)(y+2)$$"

5. Apr 7, 2012

### SammyS

Staff Emeritus
There is a typo in the second equation. It should be:
$\displaystyle y = 2\frac{1+ce^{4x}}{1-ce^{4x}}$
as O.S. showed.

6. Apr 7, 2012

### Nano-Passion

So much for being meticulous-- its a wonder how much your brain can fool you sometime.