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












0














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 differential-equations







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  • 2




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












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


















0














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 differential-equations







share|cite|improve this question




















  • 2




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












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
















0












0








0







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 differential-equations







share|cite|improve this question















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 differential-equations




calculus differential-equations






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edited Nov 29 '18 at 14:46









Davide Giraudo

125k16150261




125k16150261










asked Sep 24 '17 at 5:13









WhiteHoleWhiteHole

216




216








  • 2




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












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
















  • 2




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












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










2




2




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






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














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






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












1 Answer
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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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    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


























      0














      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
























        0












        0








        0






        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












        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

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