(2x^2 y - 2y^2 + 2xy)dx + (x^2 - 2y)dy=0
i.e. M.dx + N.dy = 0
dM/dy = 2x^2 - 4y + 2x, dN/dx = 2x (partial differentials)
Since dM/dy =/= dN/dx, then the DE is non-exact.
Try multiplying the DE by the function f() = f(x,y)
giving, f.M.dx + f.N.dy = 0
i.e. P.dx + Q.dy = 0 (P = f.M, Q = f.N)
for an exact differential, we require dP/dy = dQ/dx (partial differentials)
i.e. df/dy*M + f*dM/dy = df/dx*N + f*dN/dx
assume f() = f(x) (i.e. f is a function of x only)
then, df/dy = 0, giving
df/dx = f{dM/dy - dN/dx)} / N = f(2x^2 - 4y + 2x - 2x} / (x^2 - 2y) = f{2x^2 - 4y) / (x^2 - 2y) = 2f
i.e. df/dx = 2f
integrating,
int df/f = int 2 dx
ln(f/K) = 2x
f = Ke^(2x)
Our exact DE is now,
Ke^(2x)(2x^2 y- 2y^2 + 2xy)dx + Ke^(2x)(x^2 -2y)dy=0
e^(2x)(2x^2 y- 2y^2 + 2xy)dx + e^(2x)(x^2 -2y)dy=0
An exact differential is obtained when one takes the total derivative of a function such as U(x,y) = const, giving
dU(x,u) = dU/dx.dx + dU/dy.dy = 0, where dU/dx and dU/dy are partial differentials
Thus we have,
dU/dx = e^(2x)(2x^2 y- 2y^2 + 2xy)
integrating partially wrt x,
U(x,y) = y.e^(2x){x^2 - y} + g(y)
And,
dU/dy = e^(2x)(x^2 - y)
integrating partially wrt y,
U(x,y) = y.e^(2x){x^2 - y) + h(x)
comparison of the two solutions gives g(y) = (h(x) = 0
Our final solution is then: U(x,y) = y.e^(2x){x^2 - y) = const