How to you find the general solution of $\frac{dr}{ds}=0.05r$?

timberwuf8r
2022-09-30
Answered

How to you find the general solution of $\frac{dr}{ds}=0.05r$?

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Cooper Gillespie

Answered 2022-10-01
Author has **4** answers

We have:

$\frac{dr}{ds}=0.05r$

Which is a First Order linear separable DE. We can simply separate the variables to get

$\int \frac{1}{r}dr=\int 0.05ds$

Then integrating gives:

$\mathrm{ln}\left|r\right|=0.05s+C$

$\therefore \left|r\right|={e}^{0.05s+C}$

$\therefore \left|r\right|={e}^{0.05s}\cdot {e}^{C}$

And as $e}^{x}>0\forall x\in \mathbb{R$, and putting $A={e}^{C}$ we can write the solution as:

$r=A{e}^{0.05s}$

$\frac{dr}{ds}=0.05r$

Which is a First Order linear separable DE. We can simply separate the variables to get

$\int \frac{1}{r}dr=\int 0.05ds$

Then integrating gives:

$\mathrm{ln}\left|r\right|=0.05s+C$

$\therefore \left|r\right|={e}^{0.05s+C}$

$\therefore \left|r\right|={e}^{0.05s}\cdot {e}^{C}$

And as $e}^{x}>0\forall x\in \mathbb{R$, and putting $A={e}^{C}$ we can write the solution as:

$r=A{e}^{0.05s}$

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For a second order differential equation (many physical systems) in one variable, I know "procedures" to compute the energy. Given

${q}^{\u2033}(t)=f(q(t),{q}^{\prime}(t)),\text{}\text{}q(0)={q}_{0},\text{}\text{}{q}^{\prime}(0)={v}_{0},$

if we're lucky we can read off the related Lagrangian L, introduce $p=\frac{\mathrm{\partial}L}{\mathrm{\partial}q}$, do a Legendre transform and we got the Hamiltonian function H(q,p) for which $\frac{\text{d}}{\text{d}t}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\u27e8q(t),p(t)\u27e9$ in phase space given by a solution with initial conditions X(0), the function H(q,p) always takes the same values. It defines surfaces in phase space indexed by initial conditions X(0).

I wonder how to view this for a priori first order systems

${q}^{\prime}(t)=f(q(t)),\text{}\text{}q(0)={q}_{0},$

where I think this must be even simpler. E.g. I thing some functional

$q(t)\mapsto q(t)-\text{an integral over}\text{}f(q(t))\text{}\text{something},$

should exists which will be constant for solutions of the equation, i.e. only depend on ${q}_{0}$. For each f, the functional dependence of this "energy" on "${q}_{0}$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot{x}\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.

${q}^{\u2033}(t)=f(q(t),{q}^{\prime}(t)),\text{}\text{}q(0)={q}_{0},\text{}\text{}{q}^{\prime}(0)={v}_{0},$

if we're lucky we can read off the related Lagrangian L, introduce $p=\frac{\mathrm{\partial}L}{\mathrm{\partial}q}$, do a Legendre transform and we got the Hamiltonian function H(q,p) for which $\frac{\text{d}}{\text{d}t}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\u27e8q(t),p(t)\u27e9$ in phase space given by a solution with initial conditions X(0), the function H(q,p) always takes the same values. It defines surfaces in phase space indexed by initial conditions X(0).

I wonder how to view this for a priori first order systems

${q}^{\prime}(t)=f(q(t)),\text{}\text{}q(0)={q}_{0},$

where I think this must be even simpler. E.g. I thing some functional

$q(t)\mapsto q(t)-\text{an integral over}\text{}f(q(t))\text{}\text{something},$

should exists which will be constant for solutions of the equation, i.e. only depend on ${q}_{0}$. For each f, the functional dependence of this "energy" on "${q}_{0}$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot{x}\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.