Exact First-Order ODE
An exact equation has the form:
where the partial derivatives of f and g satisfy
Solution Procedure
The first step for solving this problem is to relate the ode to a differential in two dimensions. For a function of two variables, Z(t,y), the differential is
If we set the differential equal to 0, we have dZ=0. If the differential is 0, then the function is a constant. The solution to dZ=0 is Z(t,y)=constant. This defines a family of curves in the t-y plane.
Let's go back to the differential equation (1). If we can make the connection that f(t,y) is the t derivative of a function Z(t,y) and g(t,y) is the y derivative of Z(t,y) then the equation says dZ=0, implying Z(t,y)=constant is the solution. We need only find Z(t,y) and we are done.
Equation (2) is a necessary condition for ode (1) to be exact. Why is (2) necessary? Recall that if
then the mixed partial second derivatives satisfy
if the derivatives are continuous.
Suppose now that we know that
where f and g are given and that the necessary condition (2) is satisfied. How do we find Z(t,y)? By integration! We have
Here we have added a constant which is a function of y, since we are integrating with respect to t. We also have
Now the constant is a function of t. We find C(y) and D(t) by requiring
The final solution is
where E is a constant.
Example 1
ODE:
First, we check if the formula is exact. We have
Since the two partial derivatives are equal (and continuous) the equation is exact.
By formula (3), we have
By formula (4), we have
Comparing equations (5) and (6), we have
Choose C(y)=2y and D(t)=0. This isn't the only solution. We could choose C(y)=2y+10 and D(t)=10, for example.
The final solution to the ode is
The constant is determined by the initial condition.