[tex]\underbrace{(-84x^3y-28x^3)}_{M(x,y)}\,\mathrm dx+\underbrace{(63x^4y-6)}_{N(x,y)}=0[/tex]
This equation is not exact, since
[tex]M_y=-84x^3[/tex]
[tex]N_x=252x^3y[/tex]
So we look for an integrating factor [tex]\mu(x,y)[/tex] such that
[tex]\mu M\,\mathrm dx+\mu N\,\mathrm dy=M^*\,\mathrm dx+N^*\,\mathrm dy=0[/tex]
For this to be exact, we require
[tex]{M^*}_y={N^*}_x[/tex]
Differentiating both sides gives
[tex]\mu_yM+\mu M_y=\mu_xN+\mu N_x[/tex]
[tex]\dfrac{\mu_y}\mu M-\dfrac{\mu_x}\mu N=N_x-M_y[/tex]
When we take [tex]\mu[/tex] to be a function of either [tex]x[/tex] or [tex]y[/tex], but not both, this partial differential equation reduces to one of the separable ordinary differential equations
[tex]\dfrac{\mu_y}\mu=\dfrac{N_x-M_y}M[/tex]
[tex]\dfrac{\mu_x}\mu=-\dfrac{N_x-M_y}N[/tex]
both of which can be solved directly for [tex]\mu[/tex] provided that the result on the right hand side of either ODE is a function of either only [tex]y[/tex] or [tex]x[/tex], respectively.
The choice of which integrating factor [tex]\mu[/tex] to look for is then decided by how easily the right hand side can be taken care of. We have
[tex]\dfrac{N_x-M_y}M=\dfrac{252x^3y+84x^3}{-84x^3y-28x^3}=-3[/tex]
On the other hand, the integral resulting from an integrating factor [tex]\mu(x)[/tex] is more complicated/impossible to deal with. So, the integrating factor must be a function of [tex]y[/tex], which means [tex]\mu_x=0[/tex], and it satisfies
[tex]\dfrac{\mu_y}\mu=\dfrac1\mu\dfrac{\mathrm d\mu}{\mathrm dy}=-3\implies\displaystyle\int\frac{\mathrm d\mu}\mu=-3\int\mathrm dy[/tex]
[tex]\implies\ln\mu=-3y\implies\mu=e^{-3y}[/tex]
Distributing the integrating factor across the original ODE, we have
[tex]\underbrace{(-84x^3y-28x^3)e^{-3y}}_{M^*(x,y)}\,\mathrm dx+\underbrace{(63x^4y-6)e^{-3y}}_{N^*(x,y)}=0[/tex]
with partial derivatives
[tex]{M^*}_y=252x^3ye^{-3y}[/tex]
[tex]{N^*}_x=252x^3ye^{-3y}[/tex]
Thus the modified ODE is exact, as required. Now we try to find a solution of the form [tex]F(x,y)=C[/tex].
[tex]F_x=M^*[/tex]
[tex]F=\displaystyle\int(-84x^3y-28x^3)e^{-3y}\,\mathrm dx[/tex]
[tex]F=-7x^4(3y+1)e^{-3y}+f(y)[/tex]
[tex]F_y=N^*[/tex]
[tex]63x^4ye^{-3y}+f'(y)=(63x^4y-6)e^{-3y}[/tex]
[tex]f'(y)=-6e^{-3y}[/tex]
[tex]f(y)=2e^{-3y}+C[/tex]
Therefore the general solution is
[tex]F(x,y)=-7x^4(3y+1)e^{-3y}+2e^{-3y}=C[/tex]
