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Thread: Solve simultaneous DE

  1. #1
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    Question Solve simultaneous DE

    The DE is
    $\frac {dx}{x^2-y^2-z^2}=\frac {dy}{2xy}=\frac {dz}{2xz}$

    We can solve the 2nd and 3rd fraction to get $y=cz$.how would you get the other equation?
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  2. #2
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    Re: Solve simultaneous DE

    The solution is using
    $\frac {xdx+z (1+c^2)dz}{x^3+xz^2 (1+c^2)}=\frac {dz}{2xz}$
    This gives
    $\frac {d (x^2+z^2 (1+c^2)}{2x (x^2+z^2 (1+c^2))}=\frac {dz}{2xz} $
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  3. #3
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    Re: Solve simultaneous DE

    Yes, y= cz satisfies \frac{dy}{2xy}= \frac{dz}{2xz}

    I would prefer to write that as z= cy which also satisfies the equation. Then replacing z with cy in \frac{dx}{x^2- y^2- z^2}= \frac{dy}{2xy} we have \frac{dx}{x^2- (c^2+ 1)y^2}= \frac{dy}{2xy}. Clearing the fractions, that is the same as 2xydx= (x^2- (c^2+ 1)y^2)dy or 2xydx- (x^2- (c^2+ 1)y^2)dy= 0. That is not exact but it is not too difficult to find an integrating factor- since these are all powers of x and y, try an integrating factor of the form x^ny^m and use the exactness condition to determine that we must have n= 0, m= -2. That is, y^{-2} is an integrating factor. Multiplying by y^{-2}, we have 2xy^{-1}dx- (x^2y^{-2}- (c^2+ 1))dy= 0. That is an exact equation because (2xy^{-1})_y= -2xy^{-2}= (-x^2y^{-2}+ (c^2+1))_x.
    Thanks from sahigg
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