# Seperable Differential Equation Invovling Partial Fractions

1. Feb 4, 2012

### MathWarrior

$y' = (3+y)(1-y)$
$y(0) = 2$

Attempt to solve:
$\frac{dy}{dx}=(3+y)(1-y)$

$(3+y)(1-y)dy = dx$

$\int(3+y)(1-y)dy = \int dx$

Partial Fractions:
$\frac{A}{(3+y)} + \frac{B}{(1-y)}$

$A(1-y)+B(3+y)$

Let y= 1 or -3

A = 1/4
B = 1/4

$\frac{1}{4}\int \frac{1}{(3+y)} dy + \frac{1}{4}\int \frac{1}{(1-y)} dy$

$1/4(ln(3+y)+ln(1-y)) + C = x$

Re-arranging.
$ln(3+y)+ln(1-y) = 4x+C$

$(3+y)+(1-y) = e^{(4x+C)}$

Did I do it right so far? How do figure out y if that is the case?

2. Feb 4, 2012

### Char. Limit

You have an error here. e^(ln(3+y) + ln(1 - y)) is not equal to (3 + y) + (1 - y). Remember your properties of exponentials. From there, if you really want y as an explicit function of x, you'll need to solve the resulting quadratic.

3. Feb 4, 2012

### ehild

You have a sign error: the integral of 1/(1-y) is -ln|1-u|
Moreover: the argument of ln has to be written as |1-y| and |3+y|.

ehild

4. Feb 4, 2012

### MathWarrior

Okay I think I fixed the mistakes mentioned.. I think but I am still unsure how to do solve for y. I tried looking up properties of exponents but not sure if I did it all right.

My new equation result is:
$\frac{1}{4}(ln(3+y)-ln(1-y)+c) = x$

$(ln(3+y)-ln(1-y)) = 4x+c$

I know I have to use the exp operator but not sure how you do:

$e^{ln(3+y)-ln(1-y)}$

I know I need to use:
$\frac{a^{b}}{a^{c}} = a^{b-c}$

$\frac {e^{ln(3+y)}} {e^{ln(1-y)}} = e^{(4x+c)}$ ??

$\frac {(3+y)} {(1-y)} = e^{(4x+c)}$
How do I get y from this mess?

Last edited: Feb 4, 2012
5. Feb 4, 2012

### ehild

The absolute value signs can not be omitted. The solution of the de is $|\frac{3+y}{1−y}|=e^{4x+c}$

Use the initial condition y=2 at t=0 to find c for the case y>1.

ehild

6. Feb 4, 2012

### MathWarrior

So I substitute in y=2 and x=0?

$|\frac{3+2}{1-3}| = e^{4x+c}$

$5 = e^{4x+c}$

$ln(5) = 4x+c$

since x = 0
$c = ln(5)$ ?

7. Feb 4, 2012

### HallsofIvy

Staff Emeritus
No,
$$\left|\frac{3+2}{1- 2}\right|$$

8. Feb 4, 2012

### ehild

Yes, c=ln5. But e4x+ln5=5e4x, so you can write the solution for the given initial condition as $\frac{y+3}{y-1}=5e^{4x}$ if y>1, as |1-y|=y-1.
If you want an explicit expression for y, then multiply the equation by y-1 and isolate y.

ehild