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Differential Equations

11111 (2).jpg

I really don't understand what this means - especially the 0^+. Please help
It means as epsilon tends to zero from the right. As in epsilon gets closer and closer to zero from values greater than zero. The question is asking you to show that as epsilon approaches zero from the right the solution to the ODE is equivalent to the solution found when the epsilon in the ODE is set to zero.

How would you solve this ODE? Find a solution to the ODE and see what happens as epsilon goes to zero.

Set epsilon=0 and then solve the ODE. Compare the two solutions.
(edited 8 years ago)
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TeeEm
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Original post by ChrisHawker97
11111 (2).jpg

I really don't understand what this means - especially the 0^+. Please help


that notation bit simply means as epsilon tends to zero from the positive side (or above as it is sometimes called)
Original post by poorform
It means as epsilon tends to zero from the right. As in epsilon gets closer and closer to zero from values greater than zero. The question is asking you to show that as epsilon approaches zero from the right the solution to the ODE is equivalent to the solution found when the epsilon in the ODE is set to zero.

How would you solve this ODE? Find a solution to the ODE and see what happens as epsilon goes to zero.

Set epsilon=0 and then solve the ODE. Compare the two solutions.


Using an auxiliary equation then using the quadratic formula?

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