MHB Making up a differential equation with no real solutions

find_the_fun
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Make up a differential equation that does not possesses any real solutions.

step 1)consider the definition of solution
Any function [math]\phi[/math] defined on an interval I and possessing at least n derivatives that are continuous on I which when substituted into an nth-order ordinary differential equation reduces the equation to an identity, is said to be a solution of the equation on the interval :confused:

So it sounds to me like the task is to find a function that is not differentiable. Where I'm stuck is every function I know of is differentiable at some point and the question (I think) is asking for a DE that has no solutions over any interval.
 
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How about the square of a derivative being equated to a negative constant?
 
find_the_fun said:
... consider the definition of solution...

Any function [math]\phi[/math] defined on an interval I and possessing at least n derivatives that are continuous on I which when substituted into an nth-order ordinary differential equation reduces the equation to an identity, is said to be a solution of the equation on the interval :confused:

I'm of the opinion that the constraint of the continuity of the derivatives of the first n order may be removed and that allows the solution of a large number of pratical problems. For example let's consider the first order ODE...

$\displaystyle x^{\ '} = \mathcal {U} (t),\ x(0)=0\ (1)$

... where $\displaystyle \mathcal {U} (*)$ is the Heaviside Step Function...

Heaviside Step Function -- from Wolfram MathWorldIt is easy to verify that the solution of (1) is...

$\displaystyle x(t) = t\ \mathcal {U} (t)\ (2)$

Kind regards

$\chi$ $\sigma$
 
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I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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