prove all solutions of differential equation have 2 horizontal asymptotes












0












$begingroup$



Prove, very carefully, that the following ODE



$$ frac{ mathrm{d} y }{mathrm{d} x } = sqrt[3]{
frac{y^2+1}{x^4+1} } $$



have ${bf two}$ horizontal asymptotes.




This question was in my exam. This was my attempt:



If $y(x)$ is a solution curve, we know it will have an horizontal asymptote if
$$ lim_{x to infty} y(x) = C $$



and when the slope tends to zero. So, we want



$$ lim_{x to infty} sqrt[3]{
frac{y^2+1}{x^4+1} } = 0 $$



but this is always true so we must have horizontal asymptotes.



This seems reasonable enough but got the problem wrong and no points awarded. What did I do wrong here?






ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you show that $y^2$ grows/falls slower than $x^4$?
    $endgroup$
    – Jonas
    Jan 9 at 22:09






  • 1




    $begingroup$
    The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
    $endgroup$
    – symplectomorphic
    Jan 9 at 22:13












  • $begingroup$
    See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
    $endgroup$
    – Mattos
    Jan 9 at 23:08
















0












$begingroup$



Prove, very carefully, that the following ODE



$$ frac{ mathrm{d} y }{mathrm{d} x } = sqrt[3]{
frac{y^2+1}{x^4+1} } $$



have ${bf two}$ horizontal asymptotes.




This question was in my exam. This was my attempt:



If $y(x)$ is a solution curve, we know it will have an horizontal asymptote if
$$ lim_{x to infty} y(x) = C $$



and when the slope tends to zero. So, we want



$$ lim_{x to infty} sqrt[3]{
frac{y^2+1}{x^4+1} } = 0 $$



but this is always true so we must have horizontal asymptotes.



This seems reasonable enough but got the problem wrong and no points awarded. What did I do wrong here?






ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you show that $y^2$ grows/falls slower than $x^4$?
    $endgroup$
    – Jonas
    Jan 9 at 22:09






  • 1




    $begingroup$
    The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
    $endgroup$
    – symplectomorphic
    Jan 9 at 22:13












  • $begingroup$
    See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
    $endgroup$
    – Mattos
    Jan 9 at 23:08














0












0








0





$begingroup$



Prove, very carefully, that the following ODE



$$ frac{ mathrm{d} y }{mathrm{d} x } = sqrt[3]{
frac{y^2+1}{x^4+1} } $$



have ${bf two}$ horizontal asymptotes.




This question was in my exam. This was my attempt:



If $y(x)$ is a solution curve, we know it will have an horizontal asymptote if
$$ lim_{x to infty} y(x) = C $$



and when the slope tends to zero. So, we want



$$ lim_{x to infty} sqrt[3]{
frac{y^2+1}{x^4+1} } = 0 $$



but this is always true so we must have horizontal asymptotes.



This seems reasonable enough but got the problem wrong and no points awarded. What did I do wrong here?






ordinary-differential-equations







share|cite|improve this question









$endgroup$





Prove, very carefully, that the following ODE



$$ frac{ mathrm{d} y }{mathrm{d} x } = sqrt[3]{
frac{y^2+1}{x^4+1} } $$



have ${bf two}$ horizontal asymptotes.




This question was in my exam. This was my attempt:



If $y(x)$ is a solution curve, we know it will have an horizontal asymptote if
$$ lim_{x to infty} y(x) = C $$



and when the slope tends to zero. So, we want



$$ lim_{x to infty} sqrt[3]{
frac{y^2+1}{x^4+1} } = 0 $$



but this is always true so we must have horizontal asymptotes.



This seems reasonable enough but got the problem wrong and no points awarded. What did I do wrong here?






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










asked Jan 9 at 22:04









NeymarNeymar

355114




355114












  • $begingroup$
    Can you show that $y^2$ grows/falls slower than $x^4$?
    $endgroup$
    – Jonas
    Jan 9 at 22:09






  • 1




    $begingroup$
    The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
    $endgroup$
    – symplectomorphic
    Jan 9 at 22:13












  • $begingroup$
    See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
    $endgroup$
    – Mattos
    Jan 9 at 23:08


















  • $begingroup$
    Can you show that $y^2$ grows/falls slower than $x^4$?
    $endgroup$
    – Jonas
    Jan 9 at 22:09






  • 1




    $begingroup$
    The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
    $endgroup$
    – symplectomorphic
    Jan 9 at 22:13












  • $begingroup$
    See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
    $endgroup$
    – Mattos
    Jan 9 at 23:08
















$begingroup$
Can you show that $y^2$ grows/falls slower than $x^4$?
$endgroup$
– Jonas
Jan 9 at 22:09




$begingroup$
Can you show that $y^2$ grows/falls slower than $x^4$?
$endgroup$
– Jonas
Jan 9 at 22:09




1




1




$begingroup$
The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
$endgroup$
– symplectomorphic
Jan 9 at 22:13






$begingroup$
The first mistake is thinking that if $f(x)to C$ as $xtoinfty$ then $f'(x)to 0$. This need not be true. Consider $f(x)=frac{1}{x}sin(x^2)$, where $fto0$ but $f'$ has no limit as $xtoinfty$. The second mistake is that your calculation of $lim_{xtoinfty}frac{dy}{dx}$ unclear: $y$ is a function of $x$, so you must explain how you know $dy/dx$ tends to zero as $xtoinfty$.
$endgroup$
– symplectomorphic
Jan 9 at 22:13














$begingroup$
See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
$endgroup$
– Mattos
Jan 9 at 23:08




$begingroup$
See here for a solution to the problem (though this doesn't exactly answer what you did wrong).
$endgroup$
– Mattos
Jan 9 at 23:08










2 Answers
2






active

oldest

votes


















0












$begingroup$

You give no argument to show that



$$ lim_{x to infty} sqrt[3]{
frac{y^2+1}{x^4+1} } = 0 $$



"is always true". This is why you deserve no points.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint.



    $$
    y' = Cto y_{infty} = C x + bto C = lim_{xtoinfty}sqrt[3]{frac{(Cx+b)^2+1}{x^4+1}} = 0
    $$



    etc.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      0












      $begingroup$

      You give no argument to show that



      $$ lim_{x to infty} sqrt[3]{
      frac{y^2+1}{x^4+1} } = 0 $$



      "is always true". This is why you deserve no points.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        You give no argument to show that



        $$ lim_{x to infty} sqrt[3]{
        frac{y^2+1}{x^4+1} } = 0 $$



        "is always true". This is why you deserve no points.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You give no argument to show that



          $$ lim_{x to infty} sqrt[3]{
          frac{y^2+1}{x^4+1} } = 0 $$



          "is always true". This is why you deserve no points.






          share|cite|improve this answer









          $endgroup$



          You give no argument to show that



          $$ lim_{x to infty} sqrt[3]{
          frac{y^2+1}{x^4+1} } = 0 $$



          "is always true". This is why you deserve no points.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 22:10









          Yves DaoustYves Daoust

          125k671222




          125k671222























              0












              $begingroup$

              Hint.



              $$
              y' = Cto y_{infty} = C x + bto C = lim_{xtoinfty}sqrt[3]{frac{(Cx+b)^2+1}{x^4+1}} = 0
              $$



              etc.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint.



                $$
                y' = Cto y_{infty} = C x + bto C = lim_{xtoinfty}sqrt[3]{frac{(Cx+b)^2+1}{x^4+1}} = 0
                $$



                etc.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint.



                  $$
                  y' = Cto y_{infty} = C x + bto C = lim_{xtoinfty}sqrt[3]{frac{(Cx+b)^2+1}{x^4+1}} = 0
                  $$



                  etc.






                  share|cite|improve this answer









                  $endgroup$



                  Hint.



                  $$
                  y' = Cto y_{infty} = C x + bto C = lim_{xtoinfty}sqrt[3]{frac{(Cx+b)^2+1}{x^4+1}} = 0
                  $$



                  etc.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 11:22









                  CesareoCesareo

                  8,6243516




                  8,6243516






























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