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 .

Printable View

- Jan 14th 2009, 08:57 AMArythInitial Value Problem
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 .

- Jan 14th 2009, 09:37 AMThePerfectHacker
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 .

- Jan 14th 2009, 10:56 AMAryth
Thanks for the help. I appreciate it.