Whenever $y'$ is in $sin(y')$ or as a power, the degree of the polynomial equation is not defined, why?












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$begingroup$


In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?






calculus ordinary-differential-equations







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  • 2




    $begingroup$
    What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
    $endgroup$
    – bof
    Sep 24 '17 at 7:47












  • $begingroup$
    Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
    $endgroup$
    – WhiteHole
    Sep 24 '17 at 8:05


















0












$begingroup$


In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
    $endgroup$
    – bof
    Sep 24 '17 at 7:47












  • $begingroup$
    Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
    $endgroup$
    – WhiteHole
    Sep 24 '17 at 8:05
















0












0








0





$begingroup$


In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$




In my textbook it is written that for $y'+ sinleft(y'right)= 0$ the degree is not defined.
Is it by definition or there is some reasoning behind this?






calculus ordinary-differential-equations




calculus ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













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edited Nov 29 '18 at 14:46









Davide Giraudo

126k16150261




126k16150261










asked Sep 24 '17 at 5:13









WhiteHoleWhiteHole

216




216








  • 2




    $begingroup$
    What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
    $endgroup$
    – bof
    Sep 24 '17 at 7:47












  • $begingroup$
    Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
    $endgroup$
    – WhiteHole
    Sep 24 '17 at 8:05
















  • 2




    $begingroup$
    What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
    $endgroup$
    – bof
    Sep 24 '17 at 7:47












  • $begingroup$
    Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
    $endgroup$
    – WhiteHole
    Sep 24 '17 at 8:05










2




2




$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47






$begingroup$
What do you think the degree ought to be? What's the degree of $$2y'-frac16(y')^3+frac1{120}(y')^5-frac1{5040}(y')^7+cdots?$$
$endgroup$
– bof
Sep 24 '17 at 7:47














$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05






$begingroup$
Ohhhhhhhhh so its not possible to assign a value, then its undefined thanks sooo much
$endgroup$
– WhiteHole
Sep 24 '17 at 8:05












1 Answer
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$begingroup$

begin{array}{l}
y' + sin left( {y'} right) = 0\
{rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
{rm{Then,the first order differential equation can be written as }}\
y' + sin left( {y'} right) = 0\
y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
end{array}



Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.






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    1 Answer
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    1 Answer
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    $begingroup$

    begin{array}{l}
    y' + sin left( {y'} right) = 0\
    {rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
    sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
    {rm{Then,the first order differential equation can be written as }}\
    y' + sin left( {y'} right) = 0\
    y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
    end{array}



    Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      begin{array}{l}
      y' + sin left( {y'} right) = 0\
      {rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
      sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
      {rm{Then,the first order differential equation can be written as }}\
      y' + sin left( {y'} right) = 0\
      y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
      end{array}



      Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        begin{array}{l}
        y' + sin left( {y'} right) = 0\
        {rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
        sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
        {rm{Then,the first order differential equation can be written as }}\
        y' + sin left( {y'} right) = 0\
        y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
        end{array}



        Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.






        share|cite|improve this answer









        $endgroup$



        begin{array}{l}
        y' + sin left( {y'} right) = 0\
        {rm{Since, by Maclaurin series expansion of \}}sin left( x right){rm {is}}\
        sin left( x right) = x - frac{{{x^3}}}{{3!}} + frac{{{x^5}}}{{5!}} - frac{{{x^7}}}{{7!}} + cdots \
        {rm{Then,the first order differential equation can be written as }}\
        y' + sin left( {y'} right) = 0\
        y' + y' - frac{{{{left( {y'} right)}^3}}}{{3!}} + frac{{{{left( {y'} right)}^5}}}{{5!}} - frac{{{{left( {y'} right)}^7}}}{{7!}} + cdots = 0
        end{array}



        Then from the above equation, it is clear that there are infinite terms in the equation, so the degree of the equation can not be defined.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 17:08









        Krishna SrivastavKrishna Srivastav

        894




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