Author Topic: TUT0801  (Read 474 times)

ZYR

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TUT0801
« on: October 06, 2019, 02:11:39 AM »
   
    $1 + (\frac{x}{y} - \sin(y))y' = 0$
   
    Rewrite the equation, we have: $1dx + (\frac{x}{y} - \sin(y))dy = 0$

    Let $M  = 1$, and $N  = \frac{x}{y}- \sin(y)$

   Then $M_y = 0$, and $N_x= \frac{1}{y}$, where $M_y \neq N_x$
    So, it is not exact.

    Let $R_1 = \frac{M_y - N_x}{M} = \frac{-1}{y}$

    So $\mu = e^{-\int  R_1 dy} = e^{\int  \frac{1}{y} dy} = e^{\ln{y}} = y$

   Then multiply $\mu = y$ on both sides, we have

    $y dx + (x - y\sin(y)) dy = 0$

    and now it is exact since $M_y = 1 = N_x$

    So $\exists\ \psi(x,y)$, such that $\psi_x = M = y$

    $\psi  = \int M dx = \int y dx = yx + h(y)$

    Then, we can get $\psi_y = x + h'(y)$

   Also, we know $\psi_y = N = x - y\sin(y)$, which implies that

    $h'(y) = - y\sin(y)$

   So $h(y) = \int h'(y) dy = \int - y\sin(y) dy$
   
   By integrating by parts, we can know

      $ h(y) = y\cos{y} -\sin{y} + c$
   
  Therefore, we can have

       $\psi = yx + y\cos{y} -\sin{y} = c$