View Full Version : Solving an equation that resembles exact equation
AlonsoMcLaren
Sep27-11, 10:12 AM
If an ODE can be written in the form M(x,y)dx+N(x,y)dy=0, where M=δf/δy and N=δf/δx. Why isn't it correct to solve the ODE using the way analogous to the way solving Exact differential equations?
LCKurtz
Sep29-11, 10:07 PM
If an ODE can be written in the form M(x,y)dx+N(x,y)dy=0, where M=δf/δy and N=δf/δx. Why isn't it correct to solve the ODE using the way analogous to the way solving Exact differential equations?
It is, but I think you didn't write what you meant.
If M = ∂F/∂x and N = ∂F/∂y, then the equation is exact and F(x,y) = C is its solution. But this can't happen unless ∂N/∂x =∂M/∂y, which is the test for exactness.
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