Results 1 to 7 of 7

Math Help - Solving a simple differential equation

  1. #1
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52
    Is this correct?
    Quote Originally Posted by xterminal01 View Post
    e^-r = e^-s + C ?
    e^(-1) = e^(0) + c
    1/e = C
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member xterminal01's Avatar
    Joined
    Nov 2007
    Posts
    52
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    29
    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}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] I need help solving a simple differential equation?
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 14th 2010, 05:01 AM
  2. Replies: 2
    Last Post: March 15th 2010, 04:33 AM
  3. solving differential equations using laplace! simple problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 4th 2010, 07:46 AM
  4. simple differential equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: June 28th 2009, 05:45 PM
  5. simple differential equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 17th 2008, 02:06 AM

Search Tags


/mathhelpforum @mathhelpforum