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Math Help - separable differential equation y'=y^2 sin x

  1. #1
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    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

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

    Where c is my constant.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Craka View Post
    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

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

    Where c is my constant.
    Yup, it's correct. And I don't think you can get a more friendly thing... ^^
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  3. #3
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    awesome thanks moo
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  4. #4
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    Quote Originally Posted by Craka View Post
    I think I'm on the right track here, but am not sure if I can tidy it up any better

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

    Where c is my constant.
    Looks OK (apart from a small and inconsequential mistake that is unimportant).
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