Exact or Non-exact Differential Equations
(2x^2y - 2y^2 + 2xy)dx+(x^2 - 2y)dy=0
i.e. Mdx+Ndy=0
where
M=∂U/∂x, and N=∂U/∂y
M=2x^2y - 2y^2 + 2xy, ∂M/∂y=2x^2-4y+2x
N=x^2 - 2y, ∂N/∂x=2x
Since ∂M/∂y≠∂N/∂x, then the DE is not exact.
However, it may be made exact by multiplying it by a function of x and y, i.e. f(x,y).
Rewrite the DE as,
Pdx+Qdy=0
where, P=f*M, and Q=f*N
Then, ∂P/∂y=∂(f*M)/∂y=f_y*M + f*M_y
and, ∂Q/∂x=∂(f*N)/∂x=f_x*N + f*N_x
For an exact differential, ∂P/∂y=∂Q/∂x, i.e.
f_y*M + f*M_y=f_x*N + f*N_x
The above equation is too difficult to solve for a function of both x and y, so I will try for a solution assuming that f() is a function of x-only, or a function of y-only.
f=f(x)⇒f_y=0: ∴(f_x=f*(M_y - N_x)/N)
f=f(y)⇒f_x=0: ∴(f_y=f*(N_x - M_y )/M)
Using ∂M/∂y and ∂N/∂x from earlier,
f_x=f*(2x^2-4y+2x-2x)/(x^2-2y)=f*(2x^2-4y)/(x^2-2y)=2f
df/dx=2f
∫df/f=∫2dx
ln(f/K)=2x
f(x)=K.e^2x
f_y=f*(2x-2x^2+4y-2x)/(2x^2 y-2y^2+2xy) = -f*(2x^2-4x)/(2x^2 y-2y^2+2xy)
The above expression does not involve a function of y-only, therefore ignore it.
Our only possible solution is f(x)=K.e^2x, so multiply the original DE by f(x)=e^2x. (ignore the constant multiplier). This gives us,
e^2x (2x^2 y - 2y^2 + 2xy)dx + e^2x (x^2 - 2y)dy=0
i.e.
∂U/∂x=e^2x (2x^2 y - 2y^2 + 2xy)
and,
∂U/∂y=e^2x (x^2 - 2y)
Integrating ∂U/∂x and ∂U/∂y partially, this gives us,
U(x,y)=ye^2x (-y+x^2 ) + g(y)
U(x,y)=e^2x (x^2 y-y^2 ) + h(x))
⇒(g(y)=0 & h(x)=0)
i.e.
U(x,y)=e^2x (x^2 y-y^2 )=Const
Final solution is,
e^2x (x^2 y-y^2 )=C