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Math Help - differencial equation

  1. #1
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    differencial equation

    sd
    Last edited by emily28; July 20th 2007 at 02:50 AM.
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  2. #2
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    It's a separable ODE.

    Try it.
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  3. #3
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    ive been trying it for nearly an hour, but can only get to the stage of

    yy'/2y+3=1/x

    because i havent dont much of these type of maths. where should i go now?!
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  4. #4
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    Quote Originally Posted by emily28 View Post
    ive been trying it for nearly an hour, but can only get to the stage of

    yy'/2y+3=1/x

    because i havent dont much of these type of maths. where should i go now?!
    Actually it should be:
    \frac{yy^{\prime}}{2y + 6} = \frac{1}{x}

    \frac{y}{2y + 6} dy = \frac{1}{x}dx

    \int \frac{y}{2y + 6} dy = \int \frac{1}{x}dx

    The RHS is
    \int \frac{y}{2y + 6} dy = \frac{1}{2} \int \frac{y}{y + 3} dy = \frac{1}{2} \int \left (1 - \frac{3}{y + 3} \right ) dy

    Can you do it now?

    -Dan
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  5. #5
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    The RHS is


    how did you get the last part with 1-3/y+3?
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  6. #6
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    \frac{y}<br />
{{y + 3}} = \frac{{y + 3 - 3}}<br />
{{y + 3}} = 1 - \frac{3}<br />
{{y + 3}}
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  7. #7
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    oh right, this is slightly embarrasing becuase you are all so good at maths, but i still not sure where to go next. i should work on the LHS right?
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  8. #8
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    Show your progress
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by emily28 View Post
    oh right, this is slightly embarrasing becuase you are all so good at maths, but i still not sure where to go next. i should work on the LHS right?
    Let me give you one more hint. Look at the RHS:
    \int \frac{1}{x} dx

    I presume that this integral is giving you no problems.

    Is there any substitution you can think of that will make
    \int \frac{1}{y + 3}dy
    look similar to the x integral?

    -Dan
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  10. #10
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    well i know the intergral of 1/x is |ln|x+c so thats the right side sorted so im making progress!! (not much but hey its better than nothing right?

    as for the LHS Is there any substitution you can think of that will make

    look similar to the x integral?

    i dont understand what you mean?
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  11. #11
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    \int {\frac{1}<br />
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}<br />
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare

    Girl, it's quite dangerous getting in ODE if you don't remember how to integrate.
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  12. #12
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    [quote=Krizalid;60174] \int {\frac{1}<br />
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}<br />
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare

    i understand how to do this, this is simple intergration but dont understand how to relate it to 1/2 intergral (1-3/(y+3) dy
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  13. #13
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    Quote Originally Posted by topsquark View Post
    \frac{1}{2} \int \left (1 - \frac{3}{y + 3} \right ) dy
    \frac{1}<br />
{2}\int {\left( {1 - \frac{3}<br />
{{y + 3}}} \right)~dy} = \frac{1}<br />
{2}\left( {\int {~dy} - 3\int {\frac{1}<br />
{{y + 3}}~dy} } \right)

    I guess we're done Emily
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  14. #14
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    sorry intergration has never been my strong point! hence all the questions :P

    so if i put it all together i get

    1/2 (y-3ln|y+3|)=ln|x|+c

    so (y-3ln|y+3|)=2ln|x|+2c is my final answer?
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  15. #15
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    Quote Originally Posted by emily28 View Post
    sd
    You know you shouldn't do that?

    Luckily, topsquark quoted your post.
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