$M_y = 8xy$
$N_x = 4xy $
$\implies$ The equation is not exact
$\frac{M_y-N_x}{N} = \frac{2}{x}$
$\frac{d\mu}{dx} = \frac{2}{x}\mu$
$\mu = x^2$
The integrating factor is $\mu = x^2$.
New equation: $4x^3y^2 + 3x^2 lnx + x^2 + 2x^4yy'=0$
$M_y = 8x^3y = N_x$ (The equation is exact)
$\phi_x = M$
$\phi = x^4y^2 + x^3\ln x + h(y)$
$\phi_y = 2x^4y + h'(y) = N$
$h'(y)=0$
$h(y)=C$
Thus, $\phi= x^4y^2 + x^3\ln x = C$
For particular solution passing (1,1):
$1^41^2 + 1^2\ln 1 = C$
$C=1$
Thus, $\phi= x^4y^2 + x^2\ln x = 1$
