Is this differential equation solvable$?$

Question

If $$v=\frac{dx}{dt},a=\frac{dv}{dt}=\frac{d^2 x}{d t^2}$$ Then :$$x^2+v^2=a^2$$ Is solvable or not ?

Answer 1

From a physics perspective, the equation you have written is ill-posed, because position, velocity and acceleration have different physical dimensionality (i.e. different units) and therefore cannot be added together. Luckily, your equation is fairly easy to rewrite into a consistent form, which would read (as a first draft) $$ a^2 = \alpha v^2+\beta x^2, \tag{draft} $$ where $\alpha$ is a constant with dimensionality $[T^{-2}]$ and $\beta$ is a constant with dimensionality $[T^{-4}]$.

If you only had one such constant, then you would be able to choose a system of units (or, in other words, to reparametrize your independent variable, $t$, into a rescaled version, $t' = \gamma t$) for which the dimensionful constant would be set to unity. However, the first draft of the rewrite involves two constants, and you can't set them both to unity simultaneously without some special structure. That special structure sets a requirement of a special relationship between $\alpha$ and $\beta$, which in this case happens to be $\beta=\alpha^2$. Thus, your equation should really read $$ a^2 = \alpha v^2+\alpha^2 x^2. $$

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But, anyways, back to the land of mathematics, where dimensional analysis can be ignored. You are asking whether the differential equation $$ \left(\frac{d^2x}{dt^2}\right)^2 = \left(\frac{dx}{dt}\right)^2 +x^2 $$ is solvable. The answer is a clear "yes". To find solutions, you can simply put them into the numerical solver of your choice, which will happily churn out solutions once you give it suitable initial conditions. A cursory look at the numerical solutions indicates that they grow exponentially, which means that an exponential Ansatz is well justified.

And indeed, if you try $x(t) = e^{ct}$, then it takes only a minimal amount of algebra to find valid solutions. Those are not all the solutions, but if you want a complete set, then you need to be significantly clearer with what kind of problem you are dealing with.