Solving an equation that resembles exact equation

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SUMMARY

The discussion focuses on the conditions under which an ordinary differential equation (ODE) can be solved as an exact equation. It establishes that for an ODE in the form M(x,y)dx + N(x,y)dy = 0, where M = ∂F/∂y and N = ∂F/∂x, the equation is only exact if the condition ∂N/∂x = ∂M/∂y holds true. This condition is critical for determining the solvability of the ODE using methods applicable to exact differential equations.

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AlonsoMcLaren
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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?
 
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AlonsoMcLaren said:
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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