Differential equation with only the trivial solution

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SUMMARY

The discussion centers on the challenge of finding a differential equation (DE) whose only solution is the trivial solution y=0. Participants explore the possibility of constructing such an equation, with suggestions including the use of functions of y' that cannot equal zero. A proposed equation is f(y')y = 0, where f(y') could be a constant function like f(y')=1. However, the complexity of the problem is highlighted by the requirement for the solution to be complex-valued, leading to further considerations of functions like f(y')=(1+(y')(y'*)). Ultimately, the discussion reveals the difficulty in identifying a nontrivial DE that meets the criteria.

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  • Familiarity with complex-valued functions
  • Knowledge of Wronskian matrices and their significance in differential equations
  • Basic concepts of function manipulation in calculus
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Bipolarity
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Homework Statement


Find a differential equation with its only (complex-valued) solution being y=0

Homework Equations


The Attempt at a Solution


I believe that there is no DE having only y=0 as its solution, but frankly I am not sure if this is the case. I would like to know whether or not this is true, so that I know in which direction I can begin working my proof (or at least a hint is appreciated).

Also, something (in my mind) tells me this problem may have some connection with Wronskian matrices, but I have no clue really.

EDIT: I tried playing with some random diff Eqs, it seems that there are in fact trivial things like y+y'-y'=0, but this type of answer seems to be trivial to be of substance. Is there a DE that does not reduce to a trivial y=0?

Thanks!

BiP
 
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One approach is to think of a function of y' that cannot be zero, f(y') say, then write f(y')y = 0.
 
haruspex said:
One approach is to think of a function of y' that cannot be zero, f(y') say, then write f(y')y = 0.

Well, f(y')=1 works. Still seems like kind of a cheat. Notice they also said complex valued. So something like f(y')=1+(y')^2 won't work either. Not to say you can't cook one up. Like f(y')=(1+(y')(y'*)). I'm just wondering if there is something nontrivial here.
 
Last edited:
I was thinking of ey'y=0
 
haruspex said:
I was thinking of ey'y=0

Good work! And thanks!

BiP
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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