# 2nd Order Linear ODE-Derivation of system-issue

• williamcarter
In summary, the conversation discusses the process of combining two equations, A*(dy/dt)= -k*y eq1 and A*(dz/dt)=ky-kz eq2, to get a system that involves the second derivative of dz/dt. The steps involve rewriting the first equation in terms of z, substituting the resulting expression into the first equation, and multiplying both sides by a constant to get the desired result. The conversation also includes helpful tips and hints for solving the problem.
williamcarter

## Homework Statement

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.

Last edited:
williamcarter said:
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 ## ?

williamcarter
BvU said:
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?

You want to eliminate y, so you express y in terms of z. Only the second equation allows that.

williamcarter
BvU said:
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?

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

williamcarter
BvU said:
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.

BvU said:
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!

$$A{dy\over dt} + ky = 0 \Leftrightarrow A\left [ {A\over k}{d^2z\over dt^2} + {dz\over dt} \right ] + k \left [ {A\over k}{dz\over dt} + z \right ] = 0$$
$${A^2\over k}{d^2z\over dt^2} + A{dz\over dt} + k {A\over k}{dz\over dt} + k z = 0$$ 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.

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

Last edited:

## 1. What is a 2nd order linear ODE?

A 2nd order linear ODE (ordinary differential equation) is an equation that involves the derivative of an unknown function with respect to an independent variable, and the function itself, with the highest derivative being a 2nd order derivative. This means that the equation can be written in the form of y'' + P(x)y' + Q(x)y = R(x), where P(x), Q(x), and R(x) are functions of the independent variable x.

## 2. How do you derive a system of 2nd order linear ODEs?

To derive a system of 2nd order linear ODEs, you first need to identify the dependent variable (usually denoted as y) and the independent variable (usually denoted as x) in the given problem. Then, you need to express all the derivatives in terms of these variables. Finally, you can substitute these expressions into the equation and solve for the unknown function y.

## 3. What is the purpose of deriving a system of 2nd order linear ODEs?

The purpose of deriving a system of 2nd order linear ODEs is to model real-world phenomena and solve problems related to them. These equations are commonly used in physics, engineering, and other scientific fields to describe systems that involve acceleration, motion, and other variables that can be represented by 2nd order derivatives.

## 4. What are some applications of 2nd order linear ODEs?

2nd order linear ODEs have various applications in different fields. For example, in physics, they can be used to model the motion of a pendulum, the oscillations of a spring, or the behavior of a damped harmonic oscillator. In engineering, they can be used to analyze the behavior of electrical circuits, control systems, and mechanical systems. They are also used in economics, biology, and other fields to model various phenomena.

## 5. What are some methods for solving 2nd order linear ODEs?

There are several methods for solving 2nd order linear ODEs, including the method of undetermined coefficients, variation of parameters, and the Laplace transform method. These methods involve different techniques and are used depending on the form of the equation and the given initial or boundary conditions. It is important to choose the appropriate method to effectively solve the problem at hand.

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