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2nd Order Linear ODE-Derivation of system-issue

  • #1

Homework Statement


Captures.PNG

How exactly they combined equation1 and equation2 and got that system? I don't get that part.

Homework Equations


A*(dy/dt)= -k*y eq1
A*(dz/dt)=ky-kz eq2

The Attempt at a Solution


I tried substituting the 1st ky in the 2nd equation and then differentiating but I don't know how exactly they got that system which is:

d2z/dt2 + 2* (k/A) *dz/dt +(k2/A2)*z =0

I did not understand how they got that derivation of the system by combining equation 1 and equation 2.

I am willing to understand the concept, what are the steps, and the most important, how exactly to think to get that answer.

Any hint and tips will be much appreciated.

Thank you very much in advance.
 
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Answers and Replies

  • #2
BvU
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I tried substituting the 1st ky in the 2nd equation and then differentiating
Isn't it more logical to rewrite the first equation in terms of ##z## using ## y = {A\over k} {dz\over dt} + z ## ?
 
  • #3
Isn't it more logical to rewrite the first equation in terms of ##z## using ## y = {A\over k} {dz\over dt} + z ## ?
Thank you very much for your quick reply, it is much appreciated.
May I please know what was the logic behind getting back at y instead of working with its derivative?
 
  • #4
BvU
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You want to eliminate y, so you express y in terms of z. Only the second equation allows that.
 
  • #5
You want to eliminate y, so you express y in terms of z. Only the second equation allows that.
Thank you for your fast response.
Ok , so the 1st step is to
rewrite the first equation in terms of z
What should I do after that?

Thank you in advance.
 
  • #6
BvU
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So you get $$y = {A\over k}{dz\over dt} + z $$ which means $$
{dy\over dt} = {A\over k}{d^2z\over dt^2} + {dz\over dt}
$$ and you substitute those in the first equation
 
  • #7
So you get $$y = {A\over k}{dz\over dt} + z $$ which means $$
{dy\over dt} = {A\over k}{d^2z\over dt^2} + {dz\over dt}
$$ and you substitute those in the first equation
Thank you very much for your reply, I understood the concept and logic and it makes sense now.
 
  • #8
So you get $$y = {A\over k}{dz\over dt} + z $$ which means $$
{dy\over dt} = {A\over k}{d^2z\over dt^2} + {dz\over dt}
$$ and you substitute those in the first equation
I substituted and got like this:
A*[(A/k)*d2z/dt2+(dz/dt)]+ky=0
After that, I just did the multiplication and substituted the ky term with A*dz/dt+kz and still didn't work.

Any tips for that?
Thank you!
 
  • #9
BvU
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$$
A{dy\over dt} + ky = 0 \Leftrightarrow A\left [ {A\over k}{d^2z\over dt^2} + {dz\over dt} \right ] + k \left [
{A\over k}{dz\over dt} + z \right ] = 0
$$
$$
{A^2\over k}{d^2z\over dt^2} + A{dz\over dt} + k {A\over k}{dz\over dt} + k z = 0
$$ Multiply left and right with ##k\over A^2## (*) and bingo.

(*) the reason for this is that it is convenient to have the highest order derivative with a coefficient 1.

Any tips for that?
Yeah: don't work too hastily :smile:
 
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