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Thread: Solving a simple differential equation

  1. #1
    Junior Member xterminal01's Avatar
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    Solving a simple differential equation

    1st

    y' = x/y^2
    \int\text{ }y^2dy = \int\text{ }xdx

    y^3/3 = x^2/2 + C

    y^3 = 3x^2/2 + C


    2nd

    Find the particular solution that satisfies the initial condition.

    dr/ds = e^(r-s) when condition
    r(0) = 1

    \int\text{ }1/e^r dr = \int\text{ } e^s ds

    1/e^r = e^s + C

    1/e^1= e^0+ C which means C= 1/e
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  2. #2
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    Quote Originally Posted by xterminal01 View Post


    Find the particular solution that satisfies the initial condition.

    dr/ds = e^(r-s) when condition
    r(0) = 1

    \int\text{ }1/e^r dr = \int\text{ } e^s ds

    1/e^r = e^s + C

    1/e^1= e^0+ C which means C= 1/e
    \frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}

    gives

    \frac{dr}{e^r}= \frac{ds}{e^s}

    and

    \int e^{-r}~dr= \int e^{-s}~ds

    You got it from here?
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  3. #3
    Junior Member xterminal01's Avatar
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    e^-r = e^-s + C ?
    e^(-1) = e^(0) + c
    1/e = C




    Quote Originally Posted by pickslides View Post
    \frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}

    gives

    \frac{dr}{e^r}= \frac{ds}{e^s}

    and

    \int e^{-r}~dr= \int e^{-s}~ds

    You got it from here?
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  4. #4
    Junior Member xterminal01's Avatar
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    Is this correct?
    Quote Originally Posted by xterminal01 View Post
    e^-r = e^-s + C ?
    e^(-1) = e^(0) + c
    1/e = C
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  5. #5
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    From \int e^{-r}~dr= \int e^{-s}~ds

    I get -e^{-r}=-e^{-s}+C

    then

    e^{-r}=e^{-s}+C

    -r=\ln(e^{-s}+C)

    using r(0)=1 to solve for C

    -1=\ln(e^{-0}+C)

    e^{-1}=e^{-0}+C

    e^{-1}=1+C

    C=e^{-1}-1
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  6. #6
    Junior Member xterminal01's Avatar
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    Thanks, also Is the first eqq in my post done correctly as well?
    Quote Originally Posted by pickslides View Post
    From \int e^{-r}~dr= \int e^{-s}~ds

    I get -e^{-r}=-e^{-s}+C

    then

    e^{-r}=e^{-s}+C

    -r=\ln(e^{-s}+C)

    using r(0)=1 to solve for C

    -1=\ln(e^{-0}+C)

    e^{-1}=e^{-0}+C

    e^{-1}=1+C

    C=e^{-1}-1
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  7. #7
    Master Of Puppets
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    Quote Originally Posted by xterminal01 View Post
    1st



    y' = x/y^2
    \int\text{ }y^2dy = \int\text{ }xdx

    y^3/3 = x^2/2 + C

    y^3 = 3x^2/2 + C
    Finishing

    y^3 = \frac{3x^2}{2} + C

    y = \sqrt[3]{\frac{3x^2}{2} + C}
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