Is this a valid way of verifying solutions to a differential equation?
up vote
0
down vote
favorite
Suppose I give you a non-exact ordinary differential equation $M(x,y)dx+N(x,y)dy = 0$ and I ask you to prove whether or not some function $U(x,y)=C$ satisfies the differential equation. Is the following reasoning correct?
I differentiate $U(x,y)$ so I get $dU=A(x,y)dx+B(x,y)dy=0$, then I multiply $dU$ by some function $G(x,y)$ so that $B(x,y).G(x,y)=N(x,y)$, then if $U(x,y)$ is truly a solution to the differential equation this also must hold $M(x,y)= A(x,y)G(x,y)$, if I don't get an identity here then $U$ was not a solution in the first place.
differential-equations
edited Nov 23 at 3:22
Akash Roy
1
asked Nov 23 at 3:16
Juan123
815
add a comment |
up vote
0
down vote
favorite
Suppose I give you a non-exact ordinary differential equation $M(x,y)dx+N(x,y)dy = 0$ and I ask you to prove whether or not some function $U(x,y)=C$ satisfies the differential equation. Is the following reasoning correct?
I differentiate $U(x,y)$ so I get $dU=A(x,y)dx+B(x,y)dy=0$, then I multiply $dU$ by some function $G(x,y)$ so that $B(x,y).G(x,y)=N(x,y)$, then if $U(x,y)$ is truly a solution to the differential equation this also must hold $M(x,y)= A(x,y)G(x,y)$, if I don't get an identity here then $U$ was not a solution in the first place.
differential-equations
edited Nov 23 at 3:22
Akash Roy
1
asked Nov 23 at 3:16
Juan123
815
I think you are right.
– Nosrati
Nov 23 at 3:34
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I give you a non-exact ordinary differential equation $M(x,y)dx+N(x,y)dy = 0$ and I ask you to prove whether or not some function $U(x,y)=C$ satisfies the differential equation. Is the following reasoning correct?
I differentiate $U(x,y)$ so I get $dU=A(x,y)dx+B(x,y)dy=0$, then I multiply $dU$ by some function $G(x,y)$ so that $B(x,y).G(x,y)=N(x,y)$, then if $U(x,y)$ is truly a solution to the differential equation this also must hold $M(x,y)= A(x,y)G(x,y)$, if I don't get an identity here then $U$ was not a solution in the first place.
differential-equations
edited Nov 23 at 3:22
Akash Roy
1
asked Nov 23 at 3:16
Juan123
815
Suppose I give you a non-exact ordinary differential equation $M(x,y)dx+N(x,y)dy = 0$ and I ask you to prove whether or not some function $U(x,y)=C$ satisfies the differential equation. Is the following reasoning correct?
I differentiate $U(x,y)$ so I get $dU=A(x,y)dx+B(x,y)dy=0$, then I multiply $dU$ by some function $G(x,y)$ so that $B(x,y).G(x,y)=N(x,y)$, then if $U(x,y)$ is truly a solution to the differential equation this also must hold $M(x,y)= A(x,y)G(x,y)$, if I don't get an identity here then $U$ was not a solution in the first place.
differential-equations
differential-equations
edited Nov 23 at 3:22
Akash Roy
1
asked Nov 23 at 3:16
Juan123
815
edited Nov 23 at 3:22
Akash Roy
1
asked Nov 23 at 3:16
Juan123
815
edited Nov 23 at 3:22
Akash Roy
1
edited Nov 23 at 3:22
Akash Roy
1
edited Nov 23 at 3:22
Akash Roy
1
1
asked Nov 23 at 3:16
Juan123
815
asked Nov 23 at 3:16
Juan123
815
asked Nov 23 at 3:16
Juan123
815
815
I think you are right.
– Nosrati
Nov 23 at 3:34
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01
add a comment |
I think you are right.
– Nosrati
Nov 23 at 3:34
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01
I think you are right.
– Nosrati
Nov 23 at 3:34
I think you are right.
– Nosrati
Nov 23 at 3:34
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01
add a comment |
active
oldest
votes
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',
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%2f3009932%2fis-this-a-valid-way-of-verifying-solutions-to-a-differential-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3009932%2fis-this-a-valid-way-of-verifying-solutions-to-a-differential-equation%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
I think you are right.
– Nosrati
Nov 23 at 3:34
you can just check that $M+N frac {dU}{dx}=0$
– Isham
Nov 23 at 4:01