Are solutions to the differential equation $y' = 1 + x^2y^2$ growing?












1












$begingroup$


The task is to determine whether the statement is true or false.



Statement:



All solutions to the differential equation $y' = 1 + x^2y^2$ are growing.



My answer:



I think that all solution to the differential equation is growing due to that the derivative is $x^2$ and $y^2$ which always gives a positive number, which leads to a positive derivative --> growing function.



Am I thinking right or not?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you so much for tips!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:46






  • 1




    $begingroup$
    Your thinking is right.
    $endgroup$
    – Kemono Chen
    Dec 7 '18 at 9:47
















1












$begingroup$


The task is to determine whether the statement is true or false.



Statement:



All solutions to the differential equation $y' = 1 + x^2y^2$ are growing.



My answer:



I think that all solution to the differential equation is growing due to that the derivative is $x^2$ and $y^2$ which always gives a positive number, which leads to a positive derivative --> growing function.



Am I thinking right or not?






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you so much for tips!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:46






  • 1




    $begingroup$
    Your thinking is right.
    $endgroup$
    – Kemono Chen
    Dec 7 '18 at 9:47














1












1








1


1



$begingroup$


The task is to determine whether the statement is true or false.



Statement:



All solutions to the differential equation $y' = 1 + x^2y^2$ are growing.



My answer:



I think that all solution to the differential equation is growing due to that the derivative is $x^2$ and $y^2$ which always gives a positive number, which leads to a positive derivative --> growing function.



Am I thinking right or not?






ordinary-differential-equations







share|cite|improve this question











$endgroup$




The task is to determine whether the statement is true or false.



Statement:



All solutions to the differential equation $y' = 1 + x^2y^2$ are growing.



My answer:



I think that all solution to the differential equation is growing due to that the derivative is $x^2$ and $y^2$ which always gives a positive number, which leads to a positive derivative --> growing function.



Am I thinking right or not?






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 Dec 7 '18 at 9:46









Eevee Trainer

5,4431936




5,4431936










asked Dec 7 '18 at 9:42









nick fletchernick fletcher

61




61












  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you so much for tips!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:46






  • 1




    $begingroup$
    Your thinking is right.
    $endgroup$
    – Kemono Chen
    Dec 7 '18 at 9:47


















  • $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 9:46










  • $begingroup$
    Thank you so much for tips!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:46






  • 1




    $begingroup$
    Your thinking is right.
    $endgroup$
    – Kemono Chen
    Dec 7 '18 at 9:47
















$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 9:46




$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 9:46












$begingroup$
Thank you so much for tips!
$endgroup$
– nick fletcher
Dec 7 '18 at 9:46




$begingroup$
Thank you so much for tips!
$endgroup$
– nick fletcher
Dec 7 '18 at 9:46




1




1




$begingroup$
Your thinking is right.
$endgroup$
– Kemono Chen
Dec 7 '18 at 9:47




$begingroup$
Your thinking is right.
$endgroup$
– Kemono Chen
Dec 7 '18 at 9:47










2 Answers
2






active

oldest

votes


















2












$begingroup$

Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.



Since $x^2,y^2 geq 0$ for all $x,y$, then



$$y' = 1 +x^2y^2 geq 1 + 0 = 1 > 0$$



ergo, the solutions should be growing, as $y' > 0$ always.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much!!!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:49



















1












$begingroup$

Moreover, for $xge1$ you get
$$
y'ge1+y^2implies y(x)getan(x+c)text{ where }y(1)=tan(1+c),
$$
so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.




Other inequalities: Still for $xge1$ one gets a separable right side in
$$
y'le x^2(1+y^2)implies y(x)letan(x^3/3+c) text{ where }y(1)=tan(1/3+c)
$$
For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
$$
y'stackrel{ge}{sim} x^2y^2implies y(x)stackrel{ge}{sim}frac{y(a)}{1-y(a)(x^3-a^3)/3}
$$
so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $sqrt[3,]{3/y(a)+a^3}$.






share|cite|improve this answer











$endgroup$













    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%2f3029703%2fare-solutions-to-the-differential-equation-y-1-x2y2-growing%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.



    Since $x^2,y^2 geq 0$ for all $x,y$, then



    $$y' = 1 +x^2y^2 geq 1 + 0 = 1 > 0$$



    ergo, the solutions should be growing, as $y' > 0$ always.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – nick fletcher
      Dec 7 '18 at 9:49
















    2












    $begingroup$

    Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.



    Since $x^2,y^2 geq 0$ for all $x,y$, then



    $$y' = 1 +x^2y^2 geq 1 + 0 = 1 > 0$$



    ergo, the solutions should be growing, as $y' > 0$ always.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – nick fletcher
      Dec 7 '18 at 9:49














    2












    2








    2





    $begingroup$

    Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.



    Since $x^2,y^2 geq 0$ for all $x,y$, then



    $$y' = 1 +x^2y^2 geq 1 + 0 = 1 > 0$$



    ergo, the solutions should be growing, as $y' > 0$ always.






    share|cite|improve this answer









    $endgroup$



    Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.



    Since $x^2,y^2 geq 0$ for all $x,y$, then



    $$y' = 1 +x^2y^2 geq 1 + 0 = 1 > 0$$



    ergo, the solutions should be growing, as $y' > 0$ always.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 7 '18 at 9:48









    Eevee TrainerEevee Trainer

    5,4431936




    5,4431936












    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – nick fletcher
      Dec 7 '18 at 9:49


















    • $begingroup$
      Thank you so much!!!
      $endgroup$
      – nick fletcher
      Dec 7 '18 at 9:49
















    $begingroup$
    Thank you so much!!!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:49




    $begingroup$
    Thank you so much!!!
    $endgroup$
    – nick fletcher
    Dec 7 '18 at 9:49











    1












    $begingroup$

    Moreover, for $xge1$ you get
    $$
    y'ge1+y^2implies y(x)getan(x+c)text{ where }y(1)=tan(1+c),
    $$
    so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.




    Other inequalities: Still for $xge1$ one gets a separable right side in
    $$
    y'le x^2(1+y^2)implies y(x)letan(x^3/3+c) text{ where }y(1)=tan(1/3+c)
    $$
    For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
    $$
    y'stackrel{ge}{sim} x^2y^2implies y(x)stackrel{ge}{sim}frac{y(a)}{1-y(a)(x^3-a^3)/3}
    $$
    so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $sqrt[3,]{3/y(a)+a^3}$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Moreover, for $xge1$ you get
      $$
      y'ge1+y^2implies y(x)getan(x+c)text{ where }y(1)=tan(1+c),
      $$
      so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.




      Other inequalities: Still for $xge1$ one gets a separable right side in
      $$
      y'le x^2(1+y^2)implies y(x)letan(x^3/3+c) text{ where }y(1)=tan(1/3+c)
      $$
      For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
      $$
      y'stackrel{ge}{sim} x^2y^2implies y(x)stackrel{ge}{sim}frac{y(a)}{1-y(a)(x^3-a^3)/3}
      $$
      so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $sqrt[3,]{3/y(a)+a^3}$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Moreover, for $xge1$ you get
        $$
        y'ge1+y^2implies y(x)getan(x+c)text{ where }y(1)=tan(1+c),
        $$
        so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.




        Other inequalities: Still for $xge1$ one gets a separable right side in
        $$
        y'le x^2(1+y^2)implies y(x)letan(x^3/3+c) text{ where }y(1)=tan(1/3+c)
        $$
        For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
        $$
        y'stackrel{ge}{sim} x^2y^2implies y(x)stackrel{ge}{sim}frac{y(a)}{1-y(a)(x^3-a^3)/3}
        $$
        so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $sqrt[3,]{3/y(a)+a^3}$.






        share|cite|improve this answer











        $endgroup$



        Moreover, for $xge1$ you get
        $$
        y'ge1+y^2implies y(x)getan(x+c)text{ where }y(1)=tan(1+c),
        $$
        so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.




        Other inequalities: Still for $xge1$ one gets a separable right side in
        $$
        y'le x^2(1+y^2)implies y(x)letan(x^3/3+c) text{ where }y(1)=tan(1/3+c)
        $$
        For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
        $$
        y'stackrel{ge}{sim} x^2y^2implies y(x)stackrel{ge}{sim}frac{y(a)}{1-y(a)(x^3-a^3)/3}
        $$
        so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $sqrt[3,]{3/y(a)+a^3}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 10:47

























        answered Dec 7 '18 at 10:08









        LutzLLutzL

        57k42054




        57k42054






























            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%2f3029703%2fare-solutions-to-the-differential-equation-y-1-x2y2-growing%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more