Integrating Factor problem

[Ziwei Wang]

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

[sushengq]

M_y=2x cos⁡y-2xy sin⁡y-2y cos⁡x
N_x=4x cos⁡y-2xy sin⁡y-3y cos⁡x
M_y≠N_x

According to x=(M_y-N_x)/M
μ=e^(-∫R1dy)
(M_y-N_x)/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 φ_x=M
φ=∫Mdx= x² y²  cos⁡y-y³  sin⁡x+h(y)
φ_y=N
h^' (y)= -5y⁴
h(y)=-y⁵+C
φ=x² y²  cos⁡y-y³  sin⁡x-y⁵=C

[wenlinwang]

$$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
$$