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?
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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: $\displaystyle \frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$ ........
Originally Posted by mr fantastic y(1 - y^2) = y(1 - y)(1 + y), so you have a case for partial fraction decomposition: $\displaystyle \frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y}$ ........ Could you tell me how I solve for A, B, and C? Thanks!
Originally Posted by yellowrose Could you tell me how I solve for A, B, and C? Thanks! multiply through by $\displaystyle y(1 - y)(1 + y)$ to get $\displaystyle 1 = A(1 - y)(1 + y) + By(1 + y) + Cy(1 - y)$ .......(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
$\displaystyle \frac{1} {{y\left( {1 - y^2 } \right)}} = \frac{{\left( {1 - y^2 } \right) + y^2 }} {{y\left( {1 - y^2 } \right)}} = \frac{1} {y} + \frac{y} {{1 - y^2 }}.$ Easily integrable.
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