Whenever $y'$ is in $sin(y')$ or as a power, the degree of the polynomial equation is not defined, why?
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
1 Answer
1
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votes
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.
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 29 '18 at 17:08
Krishna SrivastavKrishna Srivastav
894
894
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add a comment |
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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