Let . Show that the function is a solution to the initial value problem , , on the interval . Note that this solution becomes unbounded as x approaches . Thus, the solution exists on the interval with , but not for larger .
If for then by the chain rule, therefore, . Therefore, solves the equation with . This solution becomes unbounded because and . Now say we want to solve with (where ). Then by the existence and uniqueness theorem it means will coincide with the solution on i.e. where . But this is a problem because becomes unbounded and so cannot be differenciable on if .