By Paul Dawkins

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Suppose that we have the following differential equation. M ( x, y ) + N ( x , y ) dy =0 dx (2) Note that it’s important that it be in this form! There must be an “= 0” on one side and the sign separating the two terms must be a “+”. Now, if there is a function somewhere out there in the world, Y(x,y), so that, Y x = M ( x, y ) Y y = N ( x, y ) and then we call the differential equation exact. In these cases we can write the differential equation as Yx + Yy dy =0 dx (3) Then using the chain rule from Calculus III we can further reduce the differential equation to the following derivative, d Y ( x, y ( x ) ) = 0 dx ( ) The (implicit) solution to an exact differential equation is then Y ( x, y ) = c (4) Well, it’s the solution provided we can find Y(x,y) anyway.

Aspx Differential Equations Example 5 Find the solution and interval of validity for the following IVP. y ( 0) = 1 3 y 3e3 xy - 1 + ( 2 ye3 xy + 3 xy 2e3 xy ) y¢ = 0 Solution Let’s identify M and N and check that it’s exact. M = 3 y 3e3 xy - 1 M y = 9 y 2e3 xy + 9 xy 3e3 xy N = 2 ye3 xy + 3 xy 2e3 xy N x = 9 y 2e3 xy + 9 xy 3e3 xy So, it’s exact. With the proper simplification integrating the second one isn’t too bad. However, the first is already set up for easy integration so let’s do that one.

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