# separable differential equation y'=y^2 sin x

• Mar 18th 2009, 01:35 AM
Craka
separable differential equation y'=y^2 sin x
I think I'm on the right track here, but am not sure if I can tidy it up any better and also if its the correct answer

$\displaystyle \begin{array}{l} y' = y^2 \sin x \\ \frac{{dy}}{{dx}} = y^2 \sin x \\ \int {\frac{{dy}}{{y^2 }} = \int {(\sin x)dx} } \\ \int {(y^{ - 2} )dy = } \int {(\sin x)dx} \\ \frac{{y^{ - 1} }}{{ - 1}} = - \cos x + c \\ \frac{{ - 1}}{y} = - \cos x + c \\ y = \frac{1}{{\cos x + c}} \\ \end{array}$

Where c is my constant.
• Mar 18th 2009, 02:16 AM
Moo
Hello,
Quote:

Originally Posted by Craka
I think I'm on the right track here, but am not sure if I can tidy it up any better and also if its the correct answer

$\displaystyle \begin{array}{l} y' = y^2 \sin x \\ \frac{{dy}}{{dx}} = y^2 \sin x \\ \int {\frac{{dy}}{{y^2 }} = \int {(\sin x)dx} } \\ \int {(y^{ - 2} )dy = } \int {(\sin x)dx} \\ \frac{{y^{ - 1} }}{{ - 1}} = - \cos x + c \\ \frac{{ - 1}}{y} = - \cos x + c \\ y = \frac{1}{{\cos x + c}} \\ \end{array}$

Where c is my constant.

Yup, it's correct. And I don't think you can get a more friendly thing... ^^
• Mar 18th 2009, 02:46 AM
Craka
awesome thanks moo
• Mar 18th 2009, 04:02 AM
mr fantastic
Quote:

Originally Posted by Craka
I think I'm on the right track here, but am not sure if I can tidy it up any better

$\displaystyle \begin{array}{l} y' = y^2 \sin x \\ \frac{{dy}}{{dx}} = y^2 \sin x \\ \int {\frac{{dy}}{{y^2 }} = \int {(\sin x)dx} } \\ \int {(y^{ - 2} )dy = } \int {(\sin x)dx} \\ \frac{{y^{ - 1} }}{{ - 1}} = - \cos x + c \\ \frac{{ - 1}}{y} = - \cos x + c \\ y = \frac{1}{{\cos x + c}} \\ \end{array}$

Where c is my constant.

Looks OK (apart from a small and inconsequential mistake that is unimportant).