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Math Help - Problem with a homogeneous equation

  1. #16
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Once you've recognized the equation as homogeneous, then either v = y/x or v = x/y will render the equation separable. How do I know this? From reading my textbook. There is a proof for this fact in my textbook.
    Thats new to me. I didnt realise that either will work for a homogenous equation. Thanks!
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  2. #17
    A Plied Mathematician
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    You're welcome. Incidentally, there is a slight preference as to which substitution to use, merely as a practical matter. Let's say you write your homogeneous equation in the form

    M(x,y) dx + N(x,y) dy = 0;

    furthermore, let's say M is simpler than N. Then the substitution x = vy will often work slightly better than the other, although you will get a separable equation either way. If N is simple than M, then the substitution y = ux will often work slightly better. See Zill's A First Course in Differential Equations, 6th Ed., p. 55.
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  3. #18
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Ackbeet View Post
    You're welcome. Incidentally, there is a slight preference as to which substitution to use, merely as a practical matter. Let's say you write your homogeneous equation in the form

    M(x,y) dx + N(x,y) dy = 0;

    furthermore, let's say M is simpler than N. Then the substitution x = vy will often work slightly better than the other, although you will get a separable equation either way. If N is simple than M, then the substitution y = ux will often work slightly better. See Zill's A First Course in Differential Equations, 6th Ed., p. 55.

    If we say that  M(x,y) dx + N(x,y) dy = 0 is homogeneous from order n then we can write:

      M(x,y) dx + N(x,y) dy = x^n \{ M_1 (\frac{y}{x})dx+N_1(\frac{y}{x})dy \}=0

    Or:

     M_1 (\frac{y}{x})dx+N_1(\frac{y}{x})dy =0

    Now, the substitution  y=vx,  dy=vdx+xdv will give:

     M_1(v)dx+N_1(v)(vdx+xdv)=0

    Or:

     (M_1(v)+vN_1(v))dx+xN_1(v)dv=0

    Or separable equation:

    \frac{dx}{x}+\frac{N_1(v)dv}{M_1(v)+vN_1(v)}=0
    Last edited by Also sprach Zarathustra; June 13th 2011 at 06:31 AM.
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