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Math Help - non exact equation

  1. #1
    MHF Contributor
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    non exact equation

    (2x-4y+6)dx+(x+y-3)dy=0
    i cant make integration cofficient
    because

    (P_y-Q_x)/P is not a function of x and so is the other one

    ??

    how to solve it
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  2. #2
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    If my memory serves me right, first you would use an affine transformation to reduce the equation down to a homogeneous equation. Then you would do a second transformation which would transform the coefficients to a single variable (Elementary Differential Equations by Rainville and Bedient covers this type of equation).
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  3. #3
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    i cant get a coefficient i cant transform it to be exact
    how to solve it
    ?

    i dont know what transformations to do
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  4. #4
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    for both (P_y-Q_x)/P and (P_y-Q_x)/Q i dont get a function of one variablr i cant make it exact
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  5. #5
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    It can not be converted to an exact equation.

    This equation is called non homogeneous first order equation with linear coefficient.

    which has the general form: (a_1x+b_1y+c_1)dx+(a_2x+b_2y+c_2)dy=0<br />

    Let L_1 : a_1x+b_1y+c_1=0 and L_2 : a_2x+b_2y+c_2=0


    Now, There are two cases:

    1) If \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} then L1 & L2 are parallel.
    Your equation does not satisfy this case. Tell me If you want to know how to deal with that case.

    2) \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}.

    First step:
    Find the intersection point between L1 & L2 by solving the following two equations:

    a_1x+b_1y+c_1=0 ... (1)

    a_2x+b_2y+c_2=0 ... (2)

    Assume the intersection point is (h,k).

    Second step:
    Substitute x=u+h and y=v+k.
    Clearly dx=du and dy=dv.
    These two substituion will transform the equation to a homogeneous equation.


    So, for your equation:
    (2x-4y+6)dx+(x+y-3)dy=0

    Clearly, \dfrac{2}{1} \neq\dfrac{-4}{1}.
    i.e. you will apply the procedure of the second case.
    Can you do that?
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