Author Topic: quiz 2 TUT 5103  (Read 313 times)

yangyiq5

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quiz 2 TUT 5103
« on: October 04, 2019, 02:01:32 PM »
Question:
$x^{2}y^{3}+x(1+y^{2}){y}'=0   $
$    \mu \left ( x,y \right )=\frac{1}{xy^{3}}$
Solution:
$M = x^{2}y^{3}$
$My = 3x^{2}y^{2}$
$N= x(1+y^{2})$
$Nx = 1+y^{2}$
$since My \neq Nx $
$this equation is not exact$
$multiple \mu (x,y) at both sides$
$x +\frac{1+y^{2}}{y^{3}}{y}' = 0$
$now My = Nx = 0, this equation is exact $
$\exists \varphi (x,y) s.t. \varphi x= M and\varphi y = N$
$\varphi = \int x dx = \frac{1}{2}x^{2} + h(y)$
$\varphi y = h'(y) =\frac{1+y^{2}}{y^{3}} = \frac{1}{y^{3}}+\frac{1}{y}$
$h(y) = -1/2y^{-2} + ln\left | y \right | + c$
$so\varphi = \frac{1}{2}x^{2} -1/2y^{-2} + ln\left | y \right | + c$
$ \frac{1}{2}x^{2} -1/2y^{-2} + ln\left | y \right | = c$