Homework Help: 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.