Differential Equations-Separation of variables

**yellowrose** · February 5th 2008, 05:44 PM

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?

**mr fantastic** · February 5th 2008, 05:48 PM

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:

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

**yellowrose** · February 5th 2008, 06:01 PM

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!

**Jhevon** · February 5th 2008, 06:20 PM

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

**Krizalid** · February 5th 2008, 06:34 PM

$\frac{1}<br /> {{y\left( {1 - y^2 } \right)}} = \frac{{\left( {1 - y^2 } \right) + y^2 }}<br /> {{y\left( {1 - y^2 } \right)}} = \frac{1}<br /> {y} + \frac{y}<br /> {{1 - y^2 }}.$

Easily integrable.

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