Mixing
Finding Orthogonal Trajectories (differential equations)
Find the set of orthogonal functions on the function
$$frac{x}{y}+frac{y}{x}=C(xy)^2$$ where C is non zero.
What I tried doing was first multiplying both sides with $xy$ to get $x^2+y^2=Cx^3y^3$ and now I derived both sides to get
$$2x+2yy'=3C(x^2y^3+x^3y^2y')$$
and by grouping the parts with $y'$ we get
$$y'(3Cx^3y^2-2y)=2x-3Cx^2y^3$$
here I try to eliminate C by putting $C=frac{x^2+y^2}{x^3y^3}$ and i get
$$y'(frac{3(x^2+y^2)x^3y^2}{x^3y^3}-2y)=2x-frac{3(x^2+y^2)x^2y^3}{x^3y^3}$$
canceling out i get
$$y'(frac{3(x^2+y^2)}{y}-2y)=2x-frac{3(x^2+y^2)}{x}$$ Multiplying by $xy$ i get
$$y'(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
Now I know I have to replace $y'=frac{dy}{dx}$with $-frac{dx}{dy}$
So now I have
$$frac{-dx}{dy}(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
and i'm not sure how to go on from here.
I would really appreciate if somebody could help me from here on out. Or show me an alternate way of doing this
differential
asked Nov 20 '18 at 13:53
westlife
868
add a comment |
Find the set of orthogonal functions on the function
$$frac{x}{y}+frac{y}{x}=C(xy)^2$$ where C is non zero.
What I tried doing was first multiplying both sides with $xy$ to get $x^2+y^2=Cx^3y^3$ and now I derived both sides to get
$$2x+2yy'=3C(x^2y^3+x^3y^2y')$$
and by grouping the parts with $y'$ we get
$$y'(3Cx^3y^2-2y)=2x-3Cx^2y^3$$
here I try to eliminate C by putting $C=frac{x^2+y^2}{x^3y^3}$ and i get
$$y'(frac{3(x^2+y^2)x^3y^2}{x^3y^3}-2y)=2x-frac{3(x^2+y^2)x^2y^3}{x^3y^3}$$
canceling out i get
$$y'(frac{3(x^2+y^2)}{y}-2y)=2x-frac{3(x^2+y^2)}{x}$$ Multiplying by $xy$ i get
$$y'(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
Now I know I have to replace $y'=frac{dy}{dx}$with $-frac{dx}{dy}$
So now I have
$$frac{-dx}{dy}(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
and i'm not sure how to go on from here.
I would really appreciate if somebody could help me from here on out. Or show me an alternate way of doing this
differential
asked Nov 20 '18 at 13:53
westlife
868
1
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19
add a comment |
Find the set of orthogonal functions on the function
$$frac{x}{y}+frac{y}{x}=C(xy)^2$$ where C is non zero.
What I tried doing was first multiplying both sides with $xy$ to get $x^2+y^2=Cx^3y^3$ and now I derived both sides to get
$$2x+2yy'=3C(x^2y^3+x^3y^2y')$$
and by grouping the parts with $y'$ we get
$$y'(3Cx^3y^2-2y)=2x-3Cx^2y^3$$
here I try to eliminate C by putting $C=frac{x^2+y^2}{x^3y^3}$ and i get
$$y'(frac{3(x^2+y^2)x^3y^2}{x^3y^3}-2y)=2x-frac{3(x^2+y^2)x^2y^3}{x^3y^3}$$
canceling out i get
$$y'(frac{3(x^2+y^2)}{y}-2y)=2x-frac{3(x^2+y^2)}{x}$$ Multiplying by $xy$ i get
$$y'(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
Now I know I have to replace $y'=frac{dy}{dx}$with $-frac{dx}{dy}$
So now I have
$$frac{-dx}{dy}(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
and i'm not sure how to go on from here.
I would really appreciate if somebody could help me from here on out. Or show me an alternate way of doing this
differential
asked Nov 20 '18 at 13:53
westlife
868
Find the set of orthogonal functions on the function
$$frac{x}{y}+frac{y}{x}=C(xy)^2$$ where C is non zero.
What I tried doing was first multiplying both sides with $xy$ to get $x^2+y^2=Cx^3y^3$ and now I derived both sides to get
$$2x+2yy'=3C(x^2y^3+x^3y^2y')$$
and by grouping the parts with $y'$ we get
$$y'(3Cx^3y^2-2y)=2x-3Cx^2y^3$$
here I try to eliminate C by putting $C=frac{x^2+y^2}{x^3y^3}$ and i get
$$y'(frac{3(x^2+y^2)x^3y^2}{x^3y^3}-2y)=2x-frac{3(x^2+y^2)x^2y^3}{x^3y^3}$$
canceling out i get
$$y'(frac{3(x^2+y^2)}{y}-2y)=2x-frac{3(x^2+y^2)}{x}$$ Multiplying by $xy$ i get
$$y'(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
Now I know I have to replace $y'=frac{dy}{dx}$with $-frac{dx}{dy}$
So now I have
$$frac{-dx}{dy}(3(x^2+y^2)x-2y^2x)=2x^2y-3(x^2+y^2)y$$
and i'm not sure how to go on from here.
I would really appreciate if somebody could help me from here on out. Or show me an alternate way of doing this
differential
differential
asked Nov 20 '18 at 13:53
westlife
868
asked Nov 20 '18 at 13:53
westlife
868
asked Nov 20 '18 at 13:53
westlife
868
asked Nov 20 '18 at 13:53
westlife
868
asked Nov 20 '18 at 13:53
westlife
868
868
1
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19
add a comment |
1
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19
1
1
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19
add a comment |
0
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',
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%2f3006328%2ffinding-orthogonal-trajectories-differential-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
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%2f3006328%2ffinding-orthogonal-trajectories-differential-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
1
This seems like a homogeneous equation . Try substituting $$x=vy$$ .
– Chinmaya mishra
Nov 20 '18 at 14:14
Thanks. Completrly forgot to check that. . .
– westlife
Nov 20 '18 at 14:19