This is a differential equation. The form in which the equation is given should lead one to suspect that it might be an exact differential equation.
Recall that for any differential equation of the form (M)dx+(N)dy=0, if My=Nx (where My is the partial derivative of M with respect to y and Nx is the partial derivative of N with respect to x), then the equation is exact.
Here we have M=y+2x2 and N=x−3y2. This gives My=1 and Nx=1. Since My=Nx, the equation is exact.
To solve an exact differential equation, integrate M with respect to x, then integrate N with respect to y, and finally combine the two results, omitting one copy of duplicate terms and setting it equal to C.
∫(M)dx=∫(y+2x2)dx=xy−2x+h(y)
∫(N)dy=∫(x−3y2)dy=xy+3y+g(x)
Where h(y) and g(x) are unknown functions of y and x, respectively. (It is technically correct to include these in the integrals. They are not, however, needed for the final answer.) We now combine the two terms. Note that if we have a term that is common to both ∫Mdx and ∫Ndy, we only add one copy. We then set the result equal to some constant C.
xy−2x+3y=C
The above is an implicit general solution to the provided differential equation.
