Results 1 to 8 of 8

Math Help - differential equation with y^2

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    5

    differential equation with y^2

    Hello.

    I'm stuck at this differential equation:
    x\cdot y'+y^{2}=1, y(1)=0

    Not even sure where to start, because everything I tried failed.

    Appreciate any help.
    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
    28
    Did you try this?

     x y'+y^{2}=1

     x y'= 1-y^{2}

     \frac{dy}{dx}= \frac{1-y^{2}}{x}

     \frac{dy}{1-y^{2}}= \frac{dx}{x}

     \frac{dy}{(1-y)(1+y)}= \frac{dx}{x}

    continue..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Quote Originally Posted by pickslides View Post
    Did you try this?

     x y'+y^{2}=1

     x y'= 1-y^{2}

     \frac{dy}{dx}= \frac{1-y^{2}}{x}

     \frac{dy}{1-y^{2}}= \frac{dx}{x}

     \frac{dy}{(1-y)(1+y)}= \frac{dx}{x}

    continue..
    Ahh thank you I'll try to continue from there. I started in a similar way but I I'm so tired I accidentally swapped the  \frac{dy}{dx} to  \frac{dx}{dy}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2010
    Posts
    5
    I got stuck again.
    I got to:
     \frac{1}{2}ln(y+1)-\frac{1}{2}ln(y-1)=lnx+C
    How can I continue from here?

    I know y(1)=0 but inserting 1 will not work here..
    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
    28
     \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1)

    factor out \frac{1}{2} then combine logs, let me know how you go.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Quote Originally Posted by pickslides View Post
     \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1)

    factor out \frac{1}{2} then combine logs, let me know how you go.
    Alright. I got here now, not sure if I made any mistakes though there are so many steps.

    \frac{y+1}{y-1}=x^{2}+e^{2C}

    y=\frac{x^{2}+e^{2C}+1}{x^{2}+e^{2C}-1}

    Is this correct? How can I use the y(1)=0 here?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2010
    Posts
    5
    Quote Originally Posted by pumbster View Post
    Alright. I got here now, not sure if I made any mistakes though there are so many steps.

    \frac{y+1}{y-1}=x^{2}+e^{2C}

    y=\frac{x^{2}+e^{2C}+1}{x^{2}+e^{2C}-1}

    Is this correct? How can I use the y(1)=0 here?
    I solved it! I realized a made an error, it should of course be:
    y=\frac{x^{2}e^{2C}+1}{x^{2}e^{2C}-1}

    And I ended up with the answer:
    y = \frac{x^2-1}{x^2+1}

    Thank you so much for all the help! Really appreciated!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Do you know if your answer is correct?

    I was thinking

     \frac{1}{2}\ln(y+1)-\frac{1}{2}\ln(y-1) = \ln x +C

     \frac{1}{2}\ln\frac{(y+1)}{(y-1)} = \ln x +\ln C

     \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln x +\ln C

     \ln\sqrt{\frac{(y+1)}{(y-1)}} = \ln (Cx)

     \sqrt{\frac{(y+1)}{(y-1)}} =  Cx

     \frac{(y+1)}{(y-1)} =  Cx^2

     \frac{2}{(y-1)}+1 =  Cx^2

    .
    .
    .
    .

     y = \frac{2}{Cx^2-1}+1

    Then use the initial condition y(1)=0 to find C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Partial Differential Equation satisfy corresponding equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 16th 2011, 07:15 PM
  2. Replies: 4
    Last Post: May 8th 2011, 12:27 PM
  3. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  4. Partial differential equation-wave equation(2)
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 6th 2009, 08:54 AM
  5. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM

Search Tags


/mathhelpforum @mathhelpforum