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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

    $\displaystyle 1st$

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

    $\displaystyle y^3/3 = x^2/2 + C$

    $\displaystyle y^3 = 3x^2/2 + C$


    $\displaystyle 2nd$

    Find the particular solution that satisfies the initial condition.

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

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

    $\displaystyle 1/e^r = e^s + C$

    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.

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

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

    $\displaystyle 1/e^r = e^s + C$

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

    gives

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

    and

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

    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
    $\displaystyle \frac{dr}{ds}= e^{r-s}= \frac{e^r}{e^s}$

    gives

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

    and

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

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

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

    then

    $\displaystyle e^{-r}=e^{-s}+C $

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

    using $\displaystyle r(0)=1$ to solve for $\displaystyle C$

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

    $\displaystyle e^{-1}=e^{-0}+C $

    $\displaystyle e^{-1}=1+C $

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

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

    then

    $\displaystyle e^{-r}=e^{-s}+C $

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

    using $\displaystyle r(0)=1$ to solve for $\displaystyle C$

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

    $\displaystyle e^{-1}=e^{-0}+C $

    $\displaystyle e^{-1}=1+C $

    $\displaystyle C=e^{-1}-1 $
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  7. #7
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    Quote Originally Posted by xterminal01 View Post
    $\displaystyle 1st$



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

    $\displaystyle y^3/3 = x^2/2 + C$

    $\displaystyle y^3 = 3x^2/2 + C$
    Finishing

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

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