Undergrad Solution to a second order differential equation

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SUMMARY

The discussion focuses on solving the second-order differential equation of the form $$4\left(\frac{d^2y}{dt^2}\right)\pm 2\left(\frac{dy}{dt}\right)^2+y=0$$, which describes damped harmonic motion with a drag force proportional to the square of velocity. The user seeks clarification on solving this equation using transcendental methods and numerical series solutions. A suggested approach involves substituting $$p = \frac{dy}{dt}$$, transforming the equation into a first-order form: $$4p \frac{dp}{dy} \pm 2 p^2 +y = 0$$, which can be solved further.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with Bernoulli's method for differential equations
  • Knowledge of transcendental functions and their applications
  • Basic concepts of damped harmonic motion and drag forces
NEXT STEPS
  • Research numerical methods for solving differential equations, specifically series solutions
  • Study the application of Bernoulli's method in solving first-order differential equations
  • Explore literature on transcendental functions and their role in differential equations
  • Learn about the mathematical modeling of damped harmonic motion
USEFUL FOR

Students and researchers in physics and engineering, particularly those focusing on dynamics and differential equations, will benefit from this discussion.

Taylor_1989
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I have currently been reading a book called 'Mathematical Methods In Physical Sciences'. Whilest I was looking at the differential section I came across a differential which I have never thought about before, which is of the form:
----------------------------------------------------------------------------------------------------------------------------------------
Directly from book

$$m\left(\frac{d^2y}{dt^2}\right)\pm l\left(\frac{dy}{dt}\right)^2+ky=0$$where the plus or minus sign must be chosen correctly at each stage of the motion
so that the retarding force opposes the motion. Let us solve the following special
case of this problem. Discuss the motion of a particle which is released from rest at
the point y = 1 when t = 0, and obeys the equation of motion

$$4\left(\frac{d^2y}{dt^2}\right)\pm 2\left(\frac{dy}{dt}\right)^2+y=0$$
-----------------------------------------------------------------------------------------------------------------------------------------And whilst I was reading how to solve it but it sort of breaks off, it starts explain the Bernoulli method then talks about the transcendental function, as stops. The actual equation in referring damped harmonic motion on a spring except that drag force is now of the form ##F=-bv^2## I assume. So I have tried to do a bit of research and not come up with much, I did think this could be done numerically by series solutions but I am interested to how this equation is solved through transcendental methods, and was wondering if some could explain the the approach on how to solve or even refer me to a bit of literature that could help me understand. This is my first time really tackling this type of differential so it a new concept and one I would like to understand.
 
Last edited:
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Is ##y## a function of ##x##? Or is it an independent variable?
 
Dewgale said:
Is ##y## a function of ##x##? Or is it an independent variable?
In there example y is the height to which the a particle from rest is released at t=0, also sorry I write the equation in the book form to make it clearer.
 
My approach would be the following: you can try making the substitution ##p = \frac{dy}{dt}##. Using chain rule, you can find that ##\frac{dp}{dt} = \frac{dp}{dy}\frac{dy}{dt} = p \frac{dp}{dy}##. This gives you

$$ 4p \frac{dp}{dy} \pm 2 p^2 +y = 0$$

This is now a first-order equation. Try and solve it from here. Hope this helps, and hopefully it's what you were looking for.

Divide through by ##4p##.
 
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Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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