Results 1 to 5 of 5

Thread: need help with two DEs

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    28

    need help with two DEs

    can't separate anything from this DE :
    $\displaystyle (xy'-y)^2=y'^2-2yy'/x+1$
    tried to group something but it didn't help
    $\displaystyle x^2y'(xy'-2y)+xy^2=y'(xy'-2y)+x$
    and the second problem is this below
    Attached Thumbnails Attached Thumbnails need help with two DEs-img_1255.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by noteiler View Post
    can't separate anything from this DE :
    $\displaystyle (xy'-y)^2=y'^2-2yy'/x+1$
    tried to group something but it didn't help
    $\displaystyle x^2y'(xy'-2y)+xy^2=y'(xy'-2y)+x$
    and the second problem is this below
    For problem 1) Multiply by $\displaystyle x^2$ so

    $\displaystyle
    x^2 \left(x y' - y\right)^2 = x^2 y'^2 - 2 x y' y + x^2
    $

    so

    $\displaystyle
    x^2 \left(x y' - y\right)^2 = \left(x y' - y\right)^2 + x^2 - y^2
    $

    or $\displaystyle x y' - y = \pm \frac{\sqrt{x^2-y^2}}{\sqrt{x^2-1}} $

    Then try $\displaystyle y = x u$

    Problem 2)

    From your answer

    $\displaystyle
    z'' = \frac{z'}{1+c_1 z'}
    $

    so

    $\displaystyle
    \left(1+c_1z'\right)z'' = z'
    $

    Integrate left and right sides once. See how that goes.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    28
    thanks alot for 1. answer, but i didn't really understand how that becomes:
    $\displaystyle
    z'' = \frac{z'}{1+c_1 z'}
    $

    i got $\displaystyle z''=p'p$ and
    $\displaystyle p = \frac{z}{1+cz}$
    but how can i get it from thease?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    Quote Originally Posted by noteiler View Post
    thanks alot for 1. answer, but i didn't really understand how that becomes:
    $\displaystyle
    z'' = \frac{z'}{1+c_1 z'}
    $

    i got $\displaystyle z''=p'p$ and
    $\displaystyle p = \frac{z}{1+cz}$
    but how can i get it from thease?
    Sorry, my bad. What I should of said is

    $\displaystyle
    y'' = \frac{y'}{1+c_1 y'}
    $

    so

    $\displaystyle
    (1+c_1 y') y'' = y'
    $

    giving, on integration

    $\displaystyle
    y' + \frac{c_1}{2} y'^2 = y + c_2
    $.

    Solve for $\displaystyle y'$ and then separate.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    28
    got it, thanks alot, Danny
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum