How to change the system in four first order equations?
$begingroup$
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$m_1frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2$
$m_2frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2$
Assume that the masses are $m_1 = 2$, $m_2 = frac{9}{4}$, and the spring constants are $k_1=1,k_2=3,k_3=frac{15}{4}$.
I substitute all value in the system above, and able to get
$frac{d^2x_1}{dt^2}=-2x_1+frac{3}{2}x_2$
$frac{d^2x_2}{dt^2}=frac{4}{3}x_1-3x_2$
Then how to convert the system about into four first order equations?
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$m_1frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2$
$m_2frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2$
Assume that the masses are $m_1 = 2$, $m_2 = frac{9}{4}$, and the spring constants are $k_1=1,k_2=3,k_3=frac{15}{4}$.
I substitute all value in the system above, and able to get
$frac{d^2x_1}{dt^2}=-2x_1+frac{3}{2}x_2$
$frac{d^2x_2}{dt^2}=frac{4}{3}x_1-3x_2$
Then how to convert the system about into four first order equations?
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$m_1frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2$
$m_2frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2$
Assume that the masses are $m_1 = 2$, $m_2 = frac{9}{4}$, and the spring constants are $k_1=1,k_2=3,k_3=frac{15}{4}$.
I substitute all value in the system above, and able to get
$frac{d^2x_1}{dt^2}=-2x_1+frac{3}{2}x_2$
$frac{d^2x_2}{dt^2}=frac{4}{3}x_1-3x_2$
Then how to convert the system about into four first order equations?
ordinary-differential-equations
$endgroup$
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$m_1frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2$
$m_2frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2$
Assume that the masses are $m_1 = 2$, $m_2 = frac{9}{4}$, and the spring constants are $k_1=1,k_2=3,k_3=frac{15}{4}$.
I substitute all value in the system above, and able to get
$frac{d^2x_1}{dt^2}=-2x_1+frac{3}{2}x_2$
$frac{d^2x_2}{dt^2}=frac{4}{3}x_1-3x_2$
Then how to convert the system about into four first order equations?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 5 '18 at 10:45
LTYLTY
285
285
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$
The system then becomes
$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$
However, I am not sure this actually helps you to solve the system. Instead, you might consider using
begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}
Do $frac{d^2}{dt^2}$ of equation $(1)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$
Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$
Then use equation $(1)$ to plug in expression for $x_2$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$
$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$
This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving
$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$
Plugging this back into equation $(1)$, we find
$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$
Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.
$endgroup$
add a comment |
$begingroup$
The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026925%2fhow-to-change-the-system-in-four-first-order-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$
The system then becomes
$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$
However, I am not sure this actually helps you to solve the system. Instead, you might consider using
begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}
Do $frac{d^2}{dt^2}$ of equation $(1)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$
Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$
Then use equation $(1)$ to plug in expression for $x_2$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$
$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$
This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving
$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$
Plugging this back into equation $(1)$, we find
$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$
Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.
$endgroup$
add a comment |
$begingroup$
To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$
The system then becomes
$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$
However, I am not sure this actually helps you to solve the system. Instead, you might consider using
begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}
Do $frac{d^2}{dt^2}$ of equation $(1)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$
Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$
Then use equation $(1)$ to plug in expression for $x_2$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$
$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$
This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving
$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$
Plugging this back into equation $(1)$, we find
$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$
Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.
$endgroup$
add a comment |
$begingroup$
To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$
The system then becomes
$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$
However, I am not sure this actually helps you to solve the system. Instead, you might consider using
begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}
Do $frac{d^2}{dt^2}$ of equation $(1)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$
Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$
Then use equation $(1)$ to plug in expression for $x_2$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$
$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$
This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving
$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$
Plugging this back into equation $(1)$, we find
$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$
Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.
$endgroup$
To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$
The system then becomes
$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$
However, I am not sure this actually helps you to solve the system. Instead, you might consider using
begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}
Do $frac{d^2}{dt^2}$ of equation $(1)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$
Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$
Then use equation $(1)$ to plug in expression for $x_2$:
$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$
$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$
This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving
$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$
Plugging this back into equation $(1)$, we find
$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$
Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.
edited Dec 5 '18 at 11:00
answered Dec 5 '18 at 10:49
glowstonetreesglowstonetrees
2,336418
2,336418
add a comment |
add a comment |
$begingroup$
The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.
$endgroup$
add a comment |
$begingroup$
The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.
$endgroup$
add a comment |
$begingroup$
The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.
$endgroup$
The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.
answered Dec 5 '18 at 11:03
LutzLLutzL
57.8k42054
57.8k42054
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026925%2fhow-to-change-the-system-in-four-first-order-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown