First you have to make the diff equation exact by multiplying by an integrating factor.
The integrating factor will be in the form: [tex]x^a y^b[/tex]
This gives:
[tex](3x^{a+2}y^b - x^a y^{b+2}) dx+ (x^{a+1} y^{b+1} - x^{a+3} y^{b-1}) dy = 0[/tex]
Now set partial derivatives equal to find a,b.
[tex]M dx + N dy = 0 \\ \\ M_y = N_x[/tex]
[tex]M_y = 3b x^{a+2} y^{b-1} -(b+2) x^a y^{b+1} \\ \\ N_x = (a+1)x^a y^{b+1} - (a+3)x^{a+2} y^{b-1}[/tex]
Setting coefficients equal we have:
[tex]a+1 = -(b+2) \\ \\ -(a+3) = 3b[/tex]
Solving this system of 2 equations yields:
[tex]a = -3, b = 0[/tex]
Now you have an exact diff equation:
[tex](3x^{-1} - x^{-3} y^2 )dx + (x^{-2}y - y^{-1}) dy = 0[/tex]
The solution is:
[tex]\int M dx= \int N dy = C[/tex]
[tex]\int M dx = \int (3 x^{-1} -x^{-3} y^2) dx = 3 ln (x) + \frac{1}{2}x^{-2} y^2 + h(y)[/tex]
[tex]\int N dy = \int (x^{-2}y - y^{-1} ) dy = \frac{1}{2}x^{-2} y^2 - ln (y) + h(x)[/tex]
h(y) and h(x) are constants in terms of the specified variable.
By equating the 2 integrals, you can see that
[tex]h(y) = -ln(y) \\ \\ h(x) = 3 ln (x)[/tex]
Finally, the general solution is:
[tex]\frac{1}{2} x^{-2} y^2 + 3ln(x) - ln(y) = C[/tex]
