Page 202 - Foundations Of Differential Calculus
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9. On Differential Equations  185
        if Z denotes any function of z, while P and Q are functions of x and y
        without involving z. In order that Z might be a function of x and y,itis
        necessary that
                                    ∂P    ∂Q
                                       =     .
                                    ∂y    ∂x
        According to this criterion, any proposed differential equation given in this
        general form can be judged to be either real or absurd. Hence, it is clear
        that the equation zdz = ydx + xdy is real. Since P = y and Q = x,we
        have
                                ∂P        ∂Q
                                    =1=      =1.
                                 ∂y       ∂x
                                     2     2
        However, the equation az dz = y dx + x dy is absurd, since
                           ∂P                 ∂Q
                              =2y      and        =2x.
                           ∂y                 ∂x
        But these are not equal.

        313. In order to investigate this criterion more completely, let P, Q, and
        R be functions of x, y, and z. Every differential equation in three variables,
        provided that it is of the first order, is of the form
                              Pdx + Qdy + Rdz =0.

        Whenever this equation is real, z will be equal to some function of x and
        y. Furthermore, its differential will have the form dz = pdx + qdy. Hence,
        if in the given equation this function of x and y is substituted for z and
        if pdx + qdy is substituted for dz, then of necessity, the result will be an
        identical equation 0 = 0. Since from the given equation we have
                                      Pdx    Qdy
                               dz = −      −     ,
                                       R      R
        if in P, Q, and R, this function is substituted for z, then necessarily we
        have
                                 P                 Q
                           p = −       and    q = −  .
                                 R                 R
        314. Since dz = pdx + qdy, from a previous demonstration we have
                                    ∂p    ∂q
                                       =    .
                                    ∂y    ∂x
        Hence, when the function in x and y is substituted for z,wehave p = −P/R
        and q = −Q/R, so that

                ∂p    −R∂P + P∂R              ∂q   −R∂Q + Q∂R
                   =                   and       =              .
                                                         2
                           2
                ∂y        R ∂y                ∂x       R ∂x
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