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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

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