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I was trying to solve a physics problem and the equation I came up with was: $$F-2kx(t)=mx''(t)$$ It is given that $x(0)=0$ and $x'(0)=0$, and my target is to find the extrema values of $x(t)$

Solving the differential equation: $$x(t)=c_2\sin{\left(\sqrt{\frac{2k}{m}}t\right)}+c_1\cos{\left(\sqrt{\frac{2k}{m}}t\right)}+\frac{F}{2k}$$ Putting in the given conditions, $c_2=0$ and $c_1=-\frac{F}{2k}$ , so: $$x(t)=\frac{F}{2k}\left(1-\cos{\left(\sqrt{\frac{2k}{m}}t\right)}\right)$$ Finding the extrema values:$$x_{min}=0$$$$x_{max}=\frac{F}{k}$$ Now my question is:

Is it possible to find the extrema values of $x(t)$ using only the given differential equation and the 2 conditions, but without actually solving the differential equation?

NOTE: The actual question that I simplified down to come up with the equation: irodov, 1.152

I used the relative acceleration of the right block with respect to the left one to create the equation.

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From

$$ F-2kx=m\ddot x\Rightarrow F \dot x-2kx\dot x=m\dot x\ddot x\Rightarrow F x-k x^2 + C = \frac m2 \dot x^2 $$

now from the initial conditions we have $C = 0$ then

$$ \frac m2\dot x^2+k x^2-Fx=0 $$

The potential energy associated to the elongation, is maximum when the kinetic fraction is null so the extrema are the solutions for

$$ k x^2-Fx = 0 $$

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  • $\begingroup$ Thanks for the answer. Could you refer to some online material which discusses this type of problems (as in solving differential equation-based problems, without actually solving the differential equation)? It would be really helpful, for I am relatively new to this topic $\endgroup$
    – Soham Saha
    Sep 9, 2023 at 9:17
  • $\begingroup$ In general, I don't think it is an easy task but in movement Mechanics, determining the energy constants helps a lot. $\endgroup$
    – Cesareo
    Sep 9, 2023 at 9:25

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