How do I find the general solution for this ode












0












$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48
















0












$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48














0












0








0





$begingroup$


$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




$$(x^2+1)(y^2-1),mathrm dx + xy ,mathrm dy = 0$$



The answer is said to be $frac{x+y}{1-xy} = tan(c)$






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 0:40







Steward

















asked Jan 3 at 0:33









StewardSteward

183




183












  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48


















  • $begingroup$
    Yes, thank you, I will edit my post..
    $endgroup$
    – Steward
    Jan 3 at 0:39






  • 1




    $begingroup$
    That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
    $endgroup$
    – DonAntonio
    Jan 3 at 0:43












  • $begingroup$
    It was provided in the textbook it self ..
    $endgroup$
    – Steward
    Jan 3 at 0:48
















$begingroup$
Yes, thank you, I will edit my post..
$endgroup$
– Steward
Jan 3 at 0:39




$begingroup$
Yes, thank you, I will edit my post..
$endgroup$
– Steward
Jan 3 at 0:39




1




1




$begingroup$
That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
$endgroup$
– DonAntonio
Jan 3 at 0:43






$begingroup$
That "answer" looks suspiciously fishy, and that $;tan c;$ looks really weird. Where did you get the answer from?
$endgroup$
– DonAntonio
Jan 3 at 0:43














$begingroup$
It was provided in the textbook it self ..
$endgroup$
– Steward
Jan 3 at 0:48




$begingroup$
It was provided in the textbook it self ..
$endgroup$
– Steward
Jan 3 at 0:48










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060141%2fhow-do-i-find-the-general-solution-for-this-ode%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









4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55
















4












$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55














4












4








4





$begingroup$

Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.






share|cite|improve this answer











$endgroup$



Hint: You can rearrange the equation to the form $$frac{x^2+1}xmathrm dx=frac y{1-y^2}mathrm dy$$Then you can integrate both sides. From here, we get
$$frac12 x^2+ln x+frac12ln(1-y^2)=C\e^{x^2}x^2(1-y^2)=e^{2C}equiv c\y^2=1-frac c{x^2 e^{x^2}}\y=frac{sqrt{x^2-ce^{-x^2}}}{x}$$This is the general solution.




Also note:
$$frac{x+y}{1-xy}=frac{x^2+sqrt{x^2-ce^{-x^2}}}{x}frac{1}{1-sqrt{x^2-ce^{-x^2}}}=frac{2x^2-ce^{-x^2}+(1+x^2)sqrt{x^2-ce^{-x^2}}}{x(1-x^2+ce^{-x^2})}$$This doesn't appear to cancel down to a constant in any obvious way.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 1:00

























answered Jan 3 at 0:37









John DoeJohn Doe

11.2k11239




11.2k11239












  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55


















  • $begingroup$
    Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
    $endgroup$
    – Steward
    Jan 3 at 0:47












  • $begingroup$
    @Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
    $endgroup$
    – DonAntonio
    Jan 3 at 0:51










  • $begingroup$
    @Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
    $endgroup$
    – John Doe
    Jan 3 at 0:51










  • $begingroup$
    In that case, what would be the correct answer ?
    $endgroup$
    – Steward
    Jan 3 at 0:54










  • $begingroup$
    @Steward see the edit.
    $endgroup$
    – John Doe
    Jan 3 at 0:55
















$begingroup$
Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
$endgroup$
– Steward
Jan 3 at 0:47






$begingroup$
Yes my problem is getting from 1/2 ln([y^2]-1) = -[(x^2)/2] -ln(x)+c to the actual desired solution
$endgroup$
– Steward
Jan 3 at 0:47














$begingroup$
@Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
$endgroup$
– DonAntonio
Jan 3 at 0:51




$begingroup$
@Steward Exactly my point in my past comment: where do you get $;tan C;$ from? What is $;C;$ , anyway?! And after applying the exponential there are still things like $;e^{x^2/2};$ ...
$endgroup$
– DonAntonio
Jan 3 at 0:51












$begingroup$
@Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
$endgroup$
– John Doe
Jan 3 at 0:51




$begingroup$
@Steward Hmm, yes I see the problem now. I don't believe that answer is correct.
$endgroup$
– John Doe
Jan 3 at 0:51












$begingroup$
In that case, what would be the correct answer ?
$endgroup$
– Steward
Jan 3 at 0:54




$begingroup$
In that case, what would be the correct answer ?
$endgroup$
– Steward
Jan 3 at 0:54












$begingroup$
@Steward see the edit.
$endgroup$
– John Doe
Jan 3 at 0:55




$begingroup$
@Steward see the edit.
$endgroup$
– John Doe
Jan 3 at 0:55


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060141%2fhow-do-i-find-the-general-solution-for-this-ode%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more