Mixing
![What was the name of the Arizal's mother? [closed]](https://lh5.googleusercontent.com/-5bl0hiLhMgs/AAAAAAAAAAI/AAAAAAAAAAA/AKxrwcZCaftjC0ZbMRFUiAcUF1101F-6Gw/s72-c-mo/photo.jpg?sz=32)
Differential equation, Second order linear or non linear.
$begingroup$
$y″ + (sin{x})y′ + (e^x)y =sinh x$ .
Could anyone tell me about linearity of this equation?
Should equation be linear as dependent variable and its derivative not multiply together? If not, what is the reason for non linearity?
ordinary-differential-equations
edited Jan 18 at 14:29
Naman Kumar
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
$endgroup$
add a comment |
$begingroup$
$y″ + (sin{x})y′ + (e^x)y =sinh x$ .
Could anyone tell me about linearity of this equation?
Should equation be linear as dependent variable and its derivative not multiply together? If not, what is the reason for non linearity?
ordinary-differential-equations
edited Jan 18 at 14:29
Naman Kumar
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
$endgroup$
add a comment |
$begingroup$
$y″ + (sin{x})y′ + (e^x)y =sinh x$ .
Could anyone tell me about linearity of this equation?
Should equation be linear as dependent variable and its derivative not multiply together? If not, what is the reason for non linearity?
ordinary-differential-equations
edited Jan 18 at 14:29
Naman Kumar
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
$endgroup$
$y″ + (sin{x})y′ + (e^x)y =sinh x$ .
Could anyone tell me about linearity of this equation?
Should equation be linear as dependent variable and its derivative not multiply together? If not, what is the reason for non linearity?
ordinary-differential-equations
ordinary-differential-equations
edited Jan 18 at 14:29
Naman Kumar
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
edited Jan 18 at 14:29
Naman Kumar
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
edited Jan 18 at 14:29
Naman Kumar
23613
edited Jan 18 at 14:29
Naman Kumar
23613
edited Jan 18 at 14:29
Naman Kumar
23613
23613
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
asked Jan 18 at 12:47
Digvijay NegiDigvijay Negi
53
53
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm assuming here that $y=y(x)$, so the derivatives of $y$ are with respect to $x$. The answer is yes this ordinary differential equation is linear. You are exactly right when you say the independent variable is not multiplying it's derivative.
Here is a neat set of notes describing linearity of ODEs, but for ease the main point is:
An ODE for $y=y(x)$ is linear if it can be written in the form:
$a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + dots + a_{1}(x)y' + a_{0}(x)y = g(x).$
Where the functions $g(x), a_{0}(x), dots , a_{n}(x)$ can be any functions of $x$. Also $y^{(n)}$ denotes the $n^{th}$ derivative of $y$.
I hope this answers your question, thanks.
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
$endgroup$
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%2f3078203%2fdifferential-equation-second-order-linear-or-non-linear%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm assuming here that $y=y(x)$, so the derivatives of $y$ are with respect to $x$. The answer is yes this ordinary differential equation is linear. You are exactly right when you say the independent variable is not multiplying it's derivative.
Here is a neat set of notes describing linearity of ODEs, but for ease the main point is:
An ODE for $y=y(x)$ is linear if it can be written in the form:
$a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + dots + a_{1}(x)y' + a_{0}(x)y = g(x).$
Where the functions $g(x), a_{0}(x), dots , a_{n}(x)$ can be any functions of $x$. Also $y^{(n)}$ denotes the $n^{th}$ derivative of $y$.
I hope this answers your question, thanks.
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
$endgroup$
add a comment |
$begingroup$
I'm assuming here that $y=y(x)$, so the derivatives of $y$ are with respect to $x$. The answer is yes this ordinary differential equation is linear. You are exactly right when you say the independent variable is not multiplying it's derivative.
Here is a neat set of notes describing linearity of ODEs, but for ease the main point is:
An ODE for $y=y(x)$ is linear if it can be written in the form:
$a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + dots + a_{1}(x)y' + a_{0}(x)y = g(x).$
Where the functions $g(x), a_{0}(x), dots , a_{n}(x)$ can be any functions of $x$. Also $y^{(n)}$ denotes the $n^{th}$ derivative of $y$.
I hope this answers your question, thanks.
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
$endgroup$
add a comment |
$begingroup$
I'm assuming here that $y=y(x)$, so the derivatives of $y$ are with respect to $x$. The answer is yes this ordinary differential equation is linear. You are exactly right when you say the independent variable is not multiplying it's derivative.
Here is a neat set of notes describing linearity of ODEs, but for ease the main point is:
An ODE for $y=y(x)$ is linear if it can be written in the form:
$a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + dots + a_{1}(x)y' + a_{0}(x)y = g(x).$
Where the functions $g(x), a_{0}(x), dots , a_{n}(x)$ can be any functions of $x$. Also $y^{(n)}$ denotes the $n^{th}$ derivative of $y$.
I hope this answers your question, thanks.
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
$endgroup$
I'm assuming here that $y=y(x)$, so the derivatives of $y$ are with respect to $x$. The answer is yes this ordinary differential equation is linear. You are exactly right when you say the independent variable is not multiplying it's derivative.
Here is a neat set of notes describing linearity of ODEs, but for ease the main point is:
An ODE for $y=y(x)$ is linear if it can be written in the form:
$a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + dots + a_{1}(x)y' + a_{0}(x)y = g(x).$
Where the functions $g(x), a_{0}(x), dots , a_{n}(x)$ can be any functions of $x$. Also $y^{(n)}$ denotes the $n^{th}$ derivative of $y$.
I hope this answers your question, thanks.
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
answered Jan 18 at 13:07
C_RichmondC_Richmond
863
863
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%2f3078203%2fdifferential-equation-second-order-linear-or-non-linear%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
