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







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  • $begingroup$
    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

5,8331936




5,8331936










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











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    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

    5,8331936




    5,8331936












    • $begingroup$
      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

        57.9k42054




        57.9k42054






























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