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A first order differential equation of the form

$ {M(x,y) dx + N(x,y) dy = 0} $

is considered exact if and only if

$ {{\partial M \over \partial y} ={\partial N \over \partial x}} $

Solving

  • Let $ {F(x,y) = \int M(x,y) dx + g(y)} $
  • Set $ {N = {\partial F \over \partial y}} $  $ {= ({\partial \over \partial y}\int M(x,y) dx) + g'(y)} $
  • Solve for, and integrate $ {g'(y)} $
  • Plug $ {g(y)} $ into $ {F(x,y)} $.

Example

$ {(6xy-y^3) dx + (4y + 3x^2-3xy^2) dy} $

$ {\Rightarrow \frac{\partial M}{\partial y}=6x-3y^2=\frac{\partial N}{\partial x}} $

$ {\Rightarrow F(x,y) = \int 6xy-y^3 dx = 3x^2y-xy^3 +g(y)} $

$ {\Rightarrow \frac{\partial F}{\partial y} = 3x^2-3xy^2 +g'(y) = N} $

$ {\Rightarrow g'(y) = 4y \Rightarrow g(y) = 2y^2 + C} $

$ {\Rightarrow F(x,y) = 3x^2y-xy^3+2y^2+C = 0} $


External References

http://en.wikipedia.org/wiki/Exact_differential_equation