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Math Help - Proving implicit equations are exact

  1. #1
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    Proving implicit equations are exact

    I have been given the equation:
    6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

    I need to
    (a) find the value of k that will make the equation exact.

    (b) Find an implicit solution when y(0)=1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sazzy3 View Post
    I have been given the equation:
    6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

    I need to
    (a) find the value of k that will make the equation exact.

    (b) Find an implicit solution when y(0)=1
    ok, it seems you are missing parentheses or something. i am simply too tired to figure out where. here is how to proceed. write your expression in the following way:

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

    here M and N are functions of x and y.

    to be exact, we required that M_y = N_x, that is, \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}

    now i'm off to bed
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  3. #3
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    continued

    i get for f(x,t)= 6xy^3 +cos (y) (when you treat y as a constant)
    the result of

    6y^3/18xy^2


    ????? is this right because i cant seem to g(x,t)= 2kxy^2- x sin (y) + y to work anywhere near that with a k value when treating x as a constant.
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  4. #4
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    Quote Originally Posted by sazzy3 View Post
    I have been given the equation:
    6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

    I need to
    (a) find the value of k that will make the equation exact.

    (b) Find an implicit solution when y(0)=1
    There is an extra "(" in your equation. If I drop it I have [tex]6xy^3+ cos(y)+ 2kx^2y^2- xsin(y)+ ydy/dx= 0.
    "Multiplying" the entire equation by dx gives
    (6xy^3+ cos(y)+ 2kx^2y^3- xsin(y))dx+ ydy= 0.

    An equation of the form M(x,y)dx+ N(x,y)dy= 0 is "exact" if and only if M_y= N_x
    Here M= 6xy^3+ cos(y)+ 2kx^2y^3- xsin(y) so [tex]M_y= 18xy^2- sin(y)+ 6kx^2y^2- xcos(y)[tex] while N= y so N_x= 0. There is NO value of k that will make that exact.

    Did you mean 6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0?
    In that case M= 6xy^3+ cos(y) so M_y= 18xy^2- sin(y) and N= 2kx^2y^2- x sin(y)+ y so N_x= 4kxy^2- sin(y). Those will be equal and the equation will be exact if 4k= 18 or if k= 9/2.

    In that case, there exist F(x,y) such that dF= 6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0. We must have F_x= 6xy^3+ cos(y). Integrating that with respect to x while holding y constant, F(x,y)= 3x^2y^3+ xcos(y)+ p(y). The "constant of integration", since we are holding y constant, may be some function of y, p(y).

    Now, differentiating that with respect to y, F_y= 9x^2y^2- xsin(y)+ p'(y) and that must be equal to 2kx^2y^2- xsin(y)+ y which, since k= 9/2 says 9x^2y^2- xsin(y)+ p'(y)= 9x^2y^2- xsin(y)+ y so that everything involving x cancels (as it must) and we have p'(y)= y. From that p(y)= \frac{1}{2}y^2+ C where C now really is a constant.

    Then F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C.

    The original equation says that dF= 0 which means F is equal to a constant: F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C= C' and, combining the fractions, x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2= C.
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