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Math Help - need help with two DEs

  1. #1
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    need help with two DEs

    can't separate anything from this DE :
    (xy'-y)^2=y'^2-2yy'/x+1
    tried to group something but it didn't help
    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  
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  2. #2
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    Quote Originally Posted by noteiler View Post
    can't separate anything from this DE :
    (xy'-y)^2=y'^2-2yy'/x+1
    tried to group something but it didn't help
    x^2y'(xy'-2y)+xy^2=y'(xy'-2y)+x
    and the second problem is this below
    For problem 1) Multiply by x^2 so

     <br />
x^2 \left(x y' - y\right)^2 = x^2 y'^2 - 2 x y' y + x^2<br />

    so

     <br />
x^2 \left(x y' - y\right)^2 = \left(x y' - y\right)^2 + x^2 - y^2<br />

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

    Then try y = x u

    Problem 2)

    From your answer

     <br />
z'' = \frac{z'}{1+c_1 z'}<br />

    so

     <br />
\left(1+c_1z'\right)z'' = z'<br />

    Integrate left and right sides once. See how that goes.
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  3. #3
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    thanks alot for 1. answer, but i didn't really understand how that becomes:
     <br />
z'' = \frac{z'}{1+c_1 z'}<br />

    i got z''=p'p and
    p = \frac{z}{1+cz}
    but how can i get it from thease?
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  4. #4
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    Quote Originally Posted by noteiler View Post
    thanks alot for 1. answer, but i didn't really understand how that becomes:
     <br />
z'' = \frac{z'}{1+c_1 z'}<br />

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

     <br />
y'' = \frac{y'}{1+c_1 y'}<br />

    so

     <br />
(1+c_1 y') y'' = y'<br />

    giving, on integration

     <br />
y' + \frac{c_1}{2} y'^2 = y + c_2<br />
.

    Solve for y' and then separate.
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  5. #5
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    got it, thanks alot, Danny
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