Seperable Differential Equation Invovling Partial Fractions

[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?

 

[Char. Limit]

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?

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.

 

[ehild]

\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

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

 

[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}

so is the answer:
\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?

 

[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

 

[MathWarrior]

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.

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) ?

 

[HallsofIvy]

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

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

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

5 = e^{4x+c}

ln(5) = 4x+c

since x = 0
c = ln(5) ?

 

[ehild]

Yes, c=ln5. But e^4x+ln5=5e^4x, 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