Author Topic: Integrating Factor problem  (Read 654 times)

Ziwei Wang

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Integrating Factor problem
« on: November 27, 2018, 12:25:36 AM »
This question is too complicated to solve.... Any suggestions on this problem?
Find the general solution of [2xycos(y)-y2cos(x)]dx+[2x2cos(y)-yx2sin(y)-3ysin(x)-5y3]dy=0
« Last Edit: November 27, 2018, 12:29:13 AM by Ziwei Wang »

sushengq

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Re: Integrating Factor problem
« Reply #1 on: November 27, 2018, 12:36:22 AM »
My=2x cos⁡y-2xy sin⁡y-2y cos⁡x
Nx=4x cos⁡y-2xy sin⁡y-3y cos⁡x
My≠Nx

According to x=(My-Nx)/M
μ=e(-∫R1dy)
(My-Nx)/M=(2x cos⁡y-2xy sin⁡y-2y cos⁡x-(4x cos⁡y-2xy sin⁡y-3y cos⁡x ))/(2x ycos⁡y-y2  cos⁡x )=-1/y
μ=e(-∫-1/y dy)=y
Multiply 2 on both side
(2xy2  cos⁡y-y3  cos⁡x  )dx+(2x2 y cos⁡y-y2 x2  sin⁡y-3y2  sin⁡x-5y4 )dy=0
Let φx=M
φ=∫Mdx= x2 y2  cos⁡y-y3  sin⁡x+h(y)
φy=N
h' (y)= -5y4
h(y)=-y5+C
φ=x2 y2  cos⁡y-y3  sin⁡x-y5=C

wenlinwang

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Re: Integrating Factor problem
« Reply #2 on: November 27, 2018, 05:10:46 PM »
$$M_y: 2xcos(y) -2xysin(y)-2ycos(x)\\
Nx:4xcos(y)-2xysin(y)-3ycos(x)\\
M_y \neq N_x\\

R = \frac{M_y -N_x}{M} =\frac {-2xcos(y) + ycos(x)}{2xycos(y) - y^2cos(x)}= \frac {-(2xcos(y)-ycos(x)}{2xycos(y)-y^2cos(x)}=\frac{-1}{y}\\

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

both side times y
$$
(2xy^2cos(y)-y^3cos(x))dx + (2x^2ycos(y) -y^2x^2sin(y)-3y^2sin(x)-5y^4)dy = 0

\phi_{(x,y)} , \phi_x = M, \phi_y = N \\

\phi = \int{M}dx = \int{2xy^2cos(y) - y^3cos(x)}dx\\
      = x^2y^2cos(y) - y^3sin(x) + h(y)
\phi_y = 2x^2ycos(y) - x^2y^2sin(y) -3y^2sin(x) + h'(y)= N\\
=2x^2ycos(y)-y^2x^2sin(y)-3y^2sin(x)-5y^4\\

 h'(y) = -5y^4\\
h(y) = -y^5\\
\phi = x^2y^2cos(y) - y^3sin(x)-y^5 = C
$$