# Finding solution without solving differential equation

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:

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

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

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

• 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 Sep 9, 2023 at 9:17
• In general, I don't think it is an easy task but in movement Mechanics, determining the energy constants helps a lot. Sep 9, 2023 at 9:25