Differential Equations-Separation of variables

• February 5th 2008, 06:44 PM
yellowrose
Differential Equations-Separation of variables
I need some help with this problem please:

solve by separation of variables.

dydx=y(1-y^2)

1/(y(1-y^2)dy=dx

How do I get it into y= form after taking the integral?
• February 5th 2008, 06:48 PM
mr fantastic
Quote:

Originally Posted by yellowrose
I need some help with this problem please:

solve by separation of variables.

dydx=y(1-y^2)

1/(y(1-y^2)dy=dx

How do I get it into y= form after taking the integral?

y(1 - y^2) = y(1 - y)(1 + y), so you have a case for partial fraction decomposition:

$\frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$ ........
• February 5th 2008, 07:01 PM
yellowrose
Quote:

Originally Posted by mr fantastic
y(1 - y^2) = y(1 - y)(1 + y), so you have a case for partial fraction decomposition:

$\frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$ ........

Could you tell me how I solve for A, B, and C? Thanks!
• February 5th 2008, 07:20 PM
Jhevon
Quote:

Originally Posted by yellowrose
Could you tell me how I solve for A, B, and C? Thanks!

multiply through by $y(1 - y)(1 + y)$ to get $1 = A(1 - y)(1 + y) + By(1 + y) + Cy(1 - y)$ .......(1)

then, plug in y = 0 into (1) and solve for A

then, plug in y = 1 into (1) and solve for B

then plug in y = -1 into (1) and solve for C
• February 5th 2008, 07:34 PM
Krizalid
$\frac{1}
{{y\left( {1 - y^2 } \right)}} = \frac{{\left( {1 - y^2 } \right) + y^2 }}
{{y\left( {1 - y^2 } \right)}} = \frac{1}
{y} + \frac{y}
{{1 - y^2 }}.$

Easily integrable.