00:01
Consider the first order differential equation, y prime plus t over t squared minus 4 times y plus the exponential of t over t minus 5.
00:10
We want to know for each of the given initial conditions, we want to determine the largest interval for t, on which the existence and uniqueness theorem guarantees the existence of a unique solution.
00:27
So first let's recall the existence and uniqueness theorem for differential equations of first order.
00:34
Let's rewrite y prime as the exponential of t over t minus 5 minus t over t squared minus 4 let's label this function f of t y so if f is continuous on a certain interval that contains our initial condition then we have a solution if in addition the partial derivative of f with respect to y is also continuous on the same interval that contained an initial condition, then the existence, the solution is unique.
01:50
So if f and f y are continuous on some interval ab that contains t knot, then the solution is unique.
02:36
So in our case, we have that f of ti is not defined if t is equal to 5, or t is equal to plus or minus 5, 2, because of these two divisions by 0 at the indicated points, which means that our intervals of existence and uniqueness, of existence and unique case of a solution, correspond to minus infinity to minus two, minus two to two, two to five, and five to plus infinity...