First order diferential equation with two integrating factors can be solved by separation of variablesM(x,y)dx+N(x,y)dy=0if f(x) and g(y) are integrating factors then the diferential equation can be solved by separation of variables.

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Kendall Wilkinson · Accepted Answer

Step 1With some careful application of definition and a lot of computation, we can prove the separability of an equation with integrating factors.Consider an integrating factor depending only on x (you gave it as f(x)). Then we can do as you said:f(x)M(x,y)dx+f(x)N(x,y)dy=0Now by definition of exactness of the above equation (which is an assumption you gave), we have that:∂(f(x)M(x,y))∂y=∂(f(x)N(x,y))∂xJust differentiate this:∂f∂yM(x,y)+f(x)∂M∂y∂f∂yM(x,y)+f(x)∂M∂yHowever, since the integrating factor f is only a variable of x, the term with ∂f∂y=0, leading us to this equation:f(x)∂M∂y=∂f∂xN(x,y)+f(x)∂N∂xNow a bit of logic here, if f is an integrating factor depending only on x, then ∂f∂x=dfdx, and actually, dfdx itself is only a function of x. This means that the RHS of the above equation is also only a function of x, i.e.q(x)=−1N(x,y)(∂N∂x−∂M∂y)Therefore, we have transformed the original equation:f(x)M(x,y)dx+f(x)N(x,y)dy=0into:dfdx=−f(x)q(x)Using an entirely similar set of steps and reasoning for the case of g(y), (i.e. g(y)M(x,y)dx+g(y)N(x,y)=0), we obtain:dgdy=−g(y)p(y)Where p(y) is defined as:p(y)=−1M(x,y)(∂N∂x−∂M∂y)This allows us to solve for integrating factors f and g directly through separation of variables, the solution of which are:f(x)=exp⁡(−∫q(x)dx)g(x)=exp⁡(−∫p(y)dy)To complete this.We can now easily find a general solution for the equation of the form:k(x,y)=C0where C0 is a constant and k(x, y) satisfies:∂k∂xdx+∂k∂ydy=0

First order diferential equation with two integrating factors

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