First order differential equations (movement of a rotary solenoid)

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Discussion Overview

The discussion revolves around solving a first-order differential equation related to the movement of a rotary solenoid. Participants explore the general solution and the method of separation of variables, addressing specific steps in the integration process and the correct interpretation of logarithmic functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents their attempt at solving the differential equation, expressing uncertainty about the correctness of their solution.
  • Another participant suggests integrating both sides of the equation, correcting the LaTeX formatting in the process.
  • A later reply clarifies that the natural logarithm should be used instead of an unspecified "in" function, indicating a misunderstanding in notation.
  • Another participant proposes a specific solution form, relating the initial condition to the solution of the differential equation.
  • One participant explains the separation of variables method as a general approach to solving such differential equations, emphasizing the need for integration constants.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the initial solution presented. There are multiple viewpoints regarding the integration steps and the interpretation of the logarithmic function.

Contextual Notes

Some participants point out potential errors in the integration process and notation, but the discussion remains focused on refining the approach rather than resolving the overall correctness of the proposed solutions.

Who May Find This Useful

Students and individuals interested in differential equations, particularly those studying first-order equations and methods of solving them, may find this discussion relevant.

Ben_Walker1978
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Homework Statement
The movement of a rotary solenoid is given by the following differential equation
Relevant Equations
$${5} \frac{\text{d}\theta}{\text{d}t}+{6}\theta=0$$
My question i am trying to solve:

Untitled.png


I have successfully done first order equations before but this one has got me a little stuck. My attempt at the general solution below:

$${5} \frac{\text{d}\theta}{\text{d}t}=-6\theta$$

$${5} \frac{\text{d}\theta}{\text{d}t} =\frac{\text{-6}\theta}{5}$$

$$d\theta = \frac{-6\theta}{5} dt $$

The integral should then = $$\theta = \frac{1}{5} in(5)$$

$$\theta = -6 \times (\frac{1}{5}) in (5) + c$$

This is my attempt an the general solution. Is this correct? Or could anyone help if i have gone wrong please?

Thank you
 
Last edited:
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¥frac{d¥theta}{¥theta}=-¥frac{6}{5} dt
You can integrate the both sides.

Edit to fix broken LaTeX:
##\frac{d \theta} \theta = -\frac 6 5 dt##
 
Last edited by a moderator:
Ben_Walker1978 said:
Homework Statement:: The movement of a rotary solenoid is given by the following differential equation
Relevant Equations:: $${5} \frac{\text{d}\theta}{\text{d}t}+{6}\theta=0$$

My question i am trying to solve:

View attachment 300338

I have successfully done first order equations before but this one has got me a little stuck. My attempt at the general solution below:

$${5} \frac{\text{d}\theta}{\text{d}t}=-6\theta$$

$${5} \frac{\text{d}\theta}{\text{d}t} =\frac{\text{-6}\theta}{5}$$

$$d\theta = \frac{-6\theta}{5} dt $$

The integral should then = $$\theta = \frac{1}{5} in(5)$$

$$\theta = -6 \times (\frac{1}{5}) in (5) + c$$
There is no "in" function. If you're reading this off your calculator it is ln (lowercase ell n), short for logarithmus naturalis, or in English, natural logarithm.
Ben_Walker1978 said:
This is my attempt an the general solution. Is this correct? Or could anyone help if i have gone wrong please?
No, your solution is not correct.
Going from ##\frac{d\theta} \theta = -6/5 dt##, you should get ##\ln|\theta| = -6/5t + C##
How do then solve for ##\theta## in that equation?
Ben_Walker1978 said:
Thank you
 
If θo it is θ(t) for t=0 then θ(t)=θo*e^(-6/5*t). How to get it?​
 
The solution to these type of differential equations is called separation of variables. Where the idea is, to separate the infinitesimal dx and dy on opposite sides of the equal signs (if possible). Then integrate with respect to each infinitesimal on each side. Do not forget your constant of integration on the right.
 
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