Solving Exact Equations

Definition A first-order ODE is said to be in differential form (DF) if it can expressed in the following way for some multivariable functions M(x, y ) and N(x, y ). M(x, y ) dx + N(x, y ) dy = 0 Every first-order ODE can be put in this form with a little algebra. Definition A DF M(x, y ) dx + N(x, y ) dy = 0 is said to be exact if there exists a function F (x, y ) such that BOTH of the conditions below are satisfied. 1. 2. ∂F ∂x ∂F ∂y = M(x, y ) = N(x, y ). Otherwise, if a function satisfying both of the above does not exist, then the DF is nonexact. How to Make the Definition of Exactness Useful Suppose we have an exact DF and we have found a function ∂F F (x, y ) such that ∂F ∂x = M(x, y ) and ∂y = N(x, y ). ∂F I In order for ∂F ∂x and ∂y to exist at a point (x0 , y0 ), the function F (x, y ) must first be continuous at (x0 , y0 ). h i ∂ ∂ ∂ ∂2F I Note that ∂x∂y = ∂x ∂y [F (x, y )] = ∂x [N(x, y )].   2 ∂ F ∂ ∂ ∂ I Note that ∂y ∂x = ∂y ∂x [F (x, y )] = ∂y [M(x, y )]. I In Calculus 3, we learned that if a function F (x, y ) is ∂2F ∂2F continuous then ∂x∂y = ∂y ∂x . I This means, ∂M ∂y = ∂N ∂x for any exact equation. Test for Exactness The DF M(x, y ) dx + N(x, y ) dy = 0 will be exact if and only if ∂N ∂M = . ∂y ∂x Example Determine if the following DFs are exact. I (2x + 1) dx + (3y − 3) dy = 0 I (3y − 3) dx + (2x + 1) dy = 0 I (6y 2 x 3 + 3x) dx + (3yx 4 − 3y ) dy = 0. I (4xy sin(2x 2 y ) + 3y ) dx + (2x 2 sin(2x 2 y ) + 3x + 2) dy = 0 Solving Exact Equations Consider the DF M(x, y ) dx + N(x, y ) dy = 0. First, determine whether the DF is exact using the Test for Exactness. If it is not exact, then you cannot solve it as an exact equation so try some other method (i.e. see if it is separable or linear). If it is exact, then we will do the following... Solving Exact Equations Since the Df is exact, there exists a function F (x, y ) such that ∂F ∂F ∂x = M and ∂y = N. Then, we can find F (x, y ) by using the fact that ∂F ∂x = M. Thus, ∂F ∂x Where G (x, y ) = = M Z = M(x, y ) dx Z ∂F dx ∂x F (x, y ) = G (x, y ) + g (y ) R M(x, y ) dx, and g (y ) is purely a function of y . Solving Exact Equations The appearance of this function g (y ) is because we have to remember in F (x, y ), M(x, y ), N(x, y ) and G (x, y ) that x and y are independent variables and so when we are taking derivatives and antiderivatives with respect to x, y is treated like a constant. Hence, g (y ) is like our ‘arbitrary constant term’ that we pick up from our antiderivative. Solving Exact Equations Hence, we have F (x, y ) = G (x, y ) + g (y ) for some arbitrary function g (y ). Now, we can use the fact that ∂F ∂y = N(x, y ) to solve for g (y ). ∂F ∂y = N(x, y ) ∂ [G (x, y ) + g (y )] = N(x, y ) ∂y ∂G + g 0 (y ) = N(x, y ) ∂y ∂G g 0 (y ) = N(x, y ) − ∂y  Z  ∂G g (y ) = N(x, y ) − dy ∂y g (y ) = H(y ) + C . Solving Exact Equations Therefore, F (x, y ) = G (x, y ) + H(y ) + C . However, this is still not the solution to the exact equation. The total differential of F (x, y ) is defined as dF = ∂F ∂F dx + dy , ∂x ∂y and is basically the same thing is differential approximation (again, a topic you may not have covered in Calculus 1) but for multivariable functions. in our case, since this is an exact equation, we have dF = M(x, y ) dx + N(x, y ) dy = 0, which says the total differential of F (x, y ) is just 0 (i.e. F (x, y ) is not changing values as x and y change. This means that F (x, y ) = C1 for some constant C1 . Solving Exact Equations Putting this together with what we had before, we now have F (x, y ) = G (x, y ) + h(y ) + C C1 = G (x, y ) + h(y ) + C C2 = G (x, y ) + h(y ) which is an implicit solution to the exact equation. Example Find an explicit solution to the IVP. dy y +x = , y (0) = 1 dx 1−x Example Find an explicit solution to the IVP. dy 3y + cos(x) =− , y (0) = 1 dx 3x + 1 Example Find an implicit solution to the IVP.

y − y sin(xy + y ) dx + x − (x + 1) sin(xy + x) dy = 0, y (1) = 0