How to solve a differential equation with a term to a power?
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
asked Mar 29 at 18:15
Ammar TarajiaAmmar Tarajia
233
233
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered Mar 29 at 18:22
mickepmickep
18.7k12351
$endgroup$
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
add a comment |
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
});
}
});
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3167430%2fhow-to-solve-a-differential-equation-with-a-term-to-a-power%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
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered Mar 29 at 18:22
mickepmickep
18.7k12351
$endgroup$
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
add a comment |
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered Mar 29 at 18:22
mickepmickep
18.7k12351
$endgroup$
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
add a comment |
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered Mar 29 at 18:22
mickepmickep
18.7k12351
$endgroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered Mar 29 at 18:22
mickepmickep
18.7k12351
answered Mar 29 at 18:22
mickepmickep
18.7k12351
answered Mar 29 at 18:22
mickepmickep
18.7k12351
answered Mar 29 at 18:22
mickepmickep
18.7k12351
18.7k12351
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
add a comment |
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
So for the substitution, could you use the chain rule? I'm fairly new to differential equations so sorry if the question sounds stupid.
$endgroup$
– Ammar Tarajia
yesterday
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
No need for chain rule here. For the substitution, I gave all details. You now need to solve the new differential equation for $v$. Then use the result to solve for $y$.
$endgroup$
– mickep
22 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
how would I go about solving such a differential equation?
$endgroup$
– Ammar Tarajia
9 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
$begingroup$
It is separable.
$endgroup$
– mickep
8 hours ago
add a comment |
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3167430%2fhow-to-solve-a-differential-equation-with-a-term-to-a-power%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
