Differential equation degree doubt












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$$frac{dy}{dx} = sin^{-1} (y)$$ The above equation is a form of $frac{dy}{dx} = f(y)$, so degree should be $1$. But if I write it as $$y = sinleft(frac{dy}{dx}right)$$ then degree is not defined as it is not a polynomial in $frac{dy}{dx}$. Please explain?






ordinary-differential-equations







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$endgroup$

















    5












    $begingroup$


    $$frac{dy}{dx} = sin^{-1} (y)$$ The above equation is a form of $frac{dy}{dx} = f(y)$, so degree should be $1$. But if I write it as $$y = sinleft(frac{dy}{dx}right)$$ then degree is not defined as it is not a polynomial in $frac{dy}{dx}$. Please explain?






    ordinary-differential-equations







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      $$frac{dy}{dx} = sin^{-1} (y)$$ The above equation is a form of $frac{dy}{dx} = f(y)$, so degree should be $1$. But if I write it as $$y = sinleft(frac{dy}{dx}right)$$ then degree is not defined as it is not a polynomial in $frac{dy}{dx}$. Please explain?






      ordinary-differential-equations







      share|cite|improve this question











      $endgroup$




      $$frac{dy}{dx} = sin^{-1} (y)$$ The above equation is a form of $frac{dy}{dx} = f(y)$, so degree should be $1$. But if I write it as $$y = sinleft(frac{dy}{dx}right)$$ then degree is not defined as it is not a polynomial in $frac{dy}{dx}$. Please explain?






      ordinary-differential-equations




      ordinary-differential-equations






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 25 '15 at 12:55









      Kamil Jarosz

      4,50331230




      4,50331230










      asked Jan 24 '15 at 14:05









      Aryaman BansalAryaman Bansal

      72110




      72110






















          3 Answers
          3






          active

          oldest

          votes


















          7





          +50







          $begingroup$

          The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:



          $$frac{dy}{dx}=sqrt[3]{x}tag1$$
          $$left(frac{dy}{dx}right)^3=xtag2$$



          The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.



          The problem is when you try to find degree of i.e. $$y=e^{y'}
          quadtext{or}quad
          y=sinleft(frac{dy}{dx}right)tag{a,b}$$
          There exist a formula that allow you define a degree of non-polynomials, namely $$deg;f(x)=lim_{xtoinfty}frac{log|f(x)|}{log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is $f(x)$ in $a$ ?
            $endgroup$
            – A---B
            Sep 23 '17 at 9:01












          • $begingroup$
            @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
            $endgroup$
            – Kamil Jarosz
            Sep 24 '17 at 8:37












          • $begingroup$
            Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
            $endgroup$
            – mathreadler
            Jul 23 '18 at 14:12





















          1












          $begingroup$

          Degree is not defined for terms of the form $sin(frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.



          In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:46










          • $begingroup$
            so give the degree of different differential equation , not the original one...please explane my confusion...
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:47






          • 1




            $begingroup$
            @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
            $endgroup$
            – TheChetan
            Dec 26 '15 at 14:02












          • $begingroup$
            please can you solve my this similar problem
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17










          • $begingroup$
            math.stackexchange.com/questions/1589320/…
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17



















          -2












          $begingroup$

          To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
          $frac{dy}{dx} = y, frac{dy}{dx} = y^2 + 1$ are of the first order.
          $ frac{d^2 y}{dx^2} = y, frac{d^2 y}{dx^2} = (frac{dy}{dx})^3 + 1$ are of second order.



          Your differential equation $frac{dy}{dx} = sin^{-1}y$ is first order even if it can be written in the equivalent form $y = sin(frac{dy}{dx}).$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            i actually want to know about the degree..
            $endgroup$
            – Aryaman Bansal
            Jan 24 '15 at 14:33










          • $begingroup$
            @AryamanBansal. First order (just as abel answered).
            $endgroup$
            – Claude Leibovici
            Jan 24 '15 at 14:35










          • $begingroup$
            @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
            $endgroup$
            – abel
            Jan 24 '15 at 14:38






          • 2




            $begingroup$
            @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
            $endgroup$
            – Singh
            Dec 22 '15 at 3:01






          • 1




            $begingroup$
            @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
            $endgroup$
            – Singh
            Dec 22 '15 at 3:42













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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7





          +50







          $begingroup$

          The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:



          $$frac{dy}{dx}=sqrt[3]{x}tag1$$
          $$left(frac{dy}{dx}right)^3=xtag2$$



          The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.



          The problem is when you try to find degree of i.e. $$y=e^{y'}
          quadtext{or}quad
          y=sinleft(frac{dy}{dx}right)tag{a,b}$$
          There exist a formula that allow you define a degree of non-polynomials, namely $$deg;f(x)=lim_{xtoinfty}frac{log|f(x)|}{log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is $f(x)$ in $a$ ?
            $endgroup$
            – A---B
            Sep 23 '17 at 9:01












          • $begingroup$
            @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
            $endgroup$
            – Kamil Jarosz
            Sep 24 '17 at 8:37












          • $begingroup$
            Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
            $endgroup$
            – mathreadler
            Jul 23 '18 at 14:12


















          7





          +50







          $begingroup$

          The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:



          $$frac{dy}{dx}=sqrt[3]{x}tag1$$
          $$left(frac{dy}{dx}right)^3=xtag2$$



          The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.



          The problem is when you try to find degree of i.e. $$y=e^{y'}
          quadtext{or}quad
          y=sinleft(frac{dy}{dx}right)tag{a,b}$$
          There exist a formula that allow you define a degree of non-polynomials, namely $$deg;f(x)=lim_{xtoinfty}frac{log|f(x)|}{log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $infty$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is $f(x)$ in $a$ ?
            $endgroup$
            – A---B
            Sep 23 '17 at 9:01












          • $begingroup$
            @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
            $endgroup$
            – Kamil Jarosz
            Sep 24 '17 at 8:37












          • $begingroup$
            Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
            $endgroup$
            – mathreadler
            Jul 23 '18 at 14:12
















          7





          +50







          7





          +50



          7




          +50



          $begingroup$

          The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:



          $$frac{dy}{dx}=sqrt[3]{x}tag1$$
          $$left(frac{dy}{dx}right)^3=xtag2$$



          The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.



          The problem is when you try to find degree of i.e. $$y=e^{y'}
          quadtext{or}quad
          y=sinleft(frac{dy}{dx}right)tag{a,b}$$
          There exist a formula that allow you define a degree of non-polynomials, namely $$deg;f(x)=lim_{xtoinfty}frac{log|f(x)|}{log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $infty$.






          share|cite|improve this answer









          $endgroup$



          The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:



          $$frac{dy}{dx}=sqrt[3]{x}tag1$$
          $$left(frac{dy}{dx}right)^3=xtag2$$



          The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.



          The problem is when you try to find degree of i.e. $$y=e^{y'}
          quadtext{or}quad
          y=sinleft(frac{dy}{dx}right)tag{a,b}$$
          There exist a formula that allow you define a degree of non-polynomials, namely $$deg;f(x)=lim_{xtoinfty}frac{log|f(x)|}{log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $infty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '15 at 13:16









          Kamil JaroszKamil Jarosz

          4,50331230




          4,50331230












          • $begingroup$
            What is $f(x)$ in $a$ ?
            $endgroup$
            – A---B
            Sep 23 '17 at 9:01












          • $begingroup$
            @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
            $endgroup$
            – Kamil Jarosz
            Sep 24 '17 at 8:37












          • $begingroup$
            Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
            $endgroup$
            – mathreadler
            Jul 23 '18 at 14:12




















          • $begingroup$
            What is $f(x)$ in $a$ ?
            $endgroup$
            – A---B
            Sep 23 '17 at 9:01












          • $begingroup$
            @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
            $endgroup$
            – Kamil Jarosz
            Sep 24 '17 at 8:37












          • $begingroup$
            Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
            $endgroup$
            – mathreadler
            Jul 23 '18 at 14:12


















          $begingroup$
          What is $f(x)$ in $a$ ?
          $endgroup$
          – A---B
          Sep 23 '17 at 9:01






          $begingroup$
          What is $f(x)$ in $a$ ?
          $endgroup$
          – A---B
          Sep 23 '17 at 9:01














          $begingroup$
          @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
          $endgroup$
          – Kamil Jarosz
          Sep 24 '17 at 8:37






          $begingroup$
          @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$).
          $endgroup$
          – Kamil Jarosz
          Sep 24 '17 at 8:37














          $begingroup$
          Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
          $endgroup$
          – mathreadler
          Jul 23 '18 at 14:12






          $begingroup$
          Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations.
          $endgroup$
          – mathreadler
          Jul 23 '18 at 14:12













          1












          $begingroup$

          Degree is not defined for terms of the form $sin(frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.



          In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:46










          • $begingroup$
            so give the degree of different differential equation , not the original one...please explane my confusion...
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:47






          • 1




            $begingroup$
            @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
            $endgroup$
            – TheChetan
            Dec 26 '15 at 14:02












          • $begingroup$
            please can you solve my this similar problem
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17










          • $begingroup$
            math.stackexchange.com/questions/1589320/…
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17
















          1












          $begingroup$

          Degree is not defined for terms of the form $sin(frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.



          In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:46










          • $begingroup$
            so give the degree of different differential equation , not the original one...please explane my confusion...
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:47






          • 1




            $begingroup$
            @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
            $endgroup$
            – TheChetan
            Dec 26 '15 at 14:02












          • $begingroup$
            please can you solve my this similar problem
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17










          • $begingroup$
            math.stackexchange.com/questions/1589320/…
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17














          1












          1








          1





          $begingroup$

          Degree is not defined for terms of the form $sin(frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.



          In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.






          share|cite|improve this answer









          $endgroup$



          Degree is not defined for terms of the form $sin(frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.



          In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '15 at 18:48









          TheChetanTheChetan

          1478




          1478












          • $begingroup$
            But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:46










          • $begingroup$
            so give the degree of different differential equation , not the original one...please explane my confusion...
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:47






          • 1




            $begingroup$
            @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
            $endgroup$
            – TheChetan
            Dec 26 '15 at 14:02












          • $begingroup$
            please can you solve my this similar problem
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17










          • $begingroup$
            math.stackexchange.com/questions/1589320/…
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17


















          • $begingroup$
            But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:46










          • $begingroup$
            so give the degree of different differential equation , not the original one...please explane my confusion...
            $endgroup$
            – neelkanth
            Dec 26 '15 at 11:47






          • 1




            $begingroup$
            @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
            $endgroup$
            – TheChetan
            Dec 26 '15 at 14:02












          • $begingroup$
            please can you solve my this similar problem
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17










          • $begingroup$
            math.stackexchange.com/questions/1589320/…
            $endgroup$
            – neelkanth
            Dec 26 '15 at 14:17
















          $begingroup$
          But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
          $endgroup$
          – neelkanth
          Dec 26 '15 at 11:46




          $begingroup$
          But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation.
          $endgroup$
          – neelkanth
          Dec 26 '15 at 11:46












          $begingroup$
          so give the degree of different differential equation , not the original one...please explane my confusion...
          $endgroup$
          – neelkanth
          Dec 26 '15 at 11:47




          $begingroup$
          so give the degree of different differential equation , not the original one...please explane my confusion...
          $endgroup$
          – neelkanth
          Dec 26 '15 at 11:47




          1




          1




          $begingroup$
          @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
          $endgroup$
          – TheChetan
          Dec 26 '15 at 14:02






          $begingroup$
          @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity).
          $endgroup$
          – TheChetan
          Dec 26 '15 at 14:02














          $begingroup$
          please can you solve my this similar problem
          $endgroup$
          – neelkanth
          Dec 26 '15 at 14:17




          $begingroup$
          please can you solve my this similar problem
          $endgroup$
          – neelkanth
          Dec 26 '15 at 14:17












          $begingroup$
          math.stackexchange.com/questions/1589320/…
          $endgroup$
          – neelkanth
          Dec 26 '15 at 14:17




          $begingroup$
          math.stackexchange.com/questions/1589320/…
          $endgroup$
          – neelkanth
          Dec 26 '15 at 14:17











          -2












          $begingroup$

          To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
          $frac{dy}{dx} = y, frac{dy}{dx} = y^2 + 1$ are of the first order.
          $ frac{d^2 y}{dx^2} = y, frac{d^2 y}{dx^2} = (frac{dy}{dx})^3 + 1$ are of second order.



          Your differential equation $frac{dy}{dx} = sin^{-1}y$ is first order even if it can be written in the equivalent form $y = sin(frac{dy}{dx}).$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            i actually want to know about the degree..
            $endgroup$
            – Aryaman Bansal
            Jan 24 '15 at 14:33










          • $begingroup$
            @AryamanBansal. First order (just as abel answered).
            $endgroup$
            – Claude Leibovici
            Jan 24 '15 at 14:35










          • $begingroup$
            @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
            $endgroup$
            – abel
            Jan 24 '15 at 14:38






          • 2




            $begingroup$
            @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
            $endgroup$
            – Singh
            Dec 22 '15 at 3:01






          • 1




            $begingroup$
            @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
            $endgroup$
            – Singh
            Dec 22 '15 at 3:42


















          -2












          $begingroup$

          To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
          $frac{dy}{dx} = y, frac{dy}{dx} = y^2 + 1$ are of the first order.
          $ frac{d^2 y}{dx^2} = y, frac{d^2 y}{dx^2} = (frac{dy}{dx})^3 + 1$ are of second order.



          Your differential equation $frac{dy}{dx} = sin^{-1}y$ is first order even if it can be written in the equivalent form $y = sin(frac{dy}{dx}).$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            i actually want to know about the degree..
            $endgroup$
            – Aryaman Bansal
            Jan 24 '15 at 14:33










          • $begingroup$
            @AryamanBansal. First order (just as abel answered).
            $endgroup$
            – Claude Leibovici
            Jan 24 '15 at 14:35










          • $begingroup$
            @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
            $endgroup$
            – abel
            Jan 24 '15 at 14:38






          • 2




            $begingroup$
            @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
            $endgroup$
            – Singh
            Dec 22 '15 at 3:01






          • 1




            $begingroup$
            @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
            $endgroup$
            – Singh
            Dec 22 '15 at 3:42
















          -2












          -2








          -2





          $begingroup$

          To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
          $frac{dy}{dx} = y, frac{dy}{dx} = y^2 + 1$ are of the first order.
          $ frac{d^2 y}{dx^2} = y, frac{d^2 y}{dx^2} = (frac{dy}{dx})^3 + 1$ are of second order.



          Your differential equation $frac{dy}{dx} = sin^{-1}y$ is first order even if it can be written in the equivalent form $y = sin(frac{dy}{dx}).$






          share|cite|improve this answer











          $endgroup$



          To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example,
          $frac{dy}{dx} = y, frac{dy}{dx} = y^2 + 1$ are of the first order.
          $ frac{d^2 y}{dx^2} = y, frac{d^2 y}{dx^2} = (frac{dy}{dx})^3 + 1$ are of second order.



          Your differential equation $frac{dy}{dx} = sin^{-1}y$ is first order even if it can be written in the equivalent form $y = sin(frac{dy}{dx}).$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited May 4 '15 at 7:10









          Aryaman Bansal

          72110




          72110










          answered Jan 24 '15 at 14:19









          abelabel

          26.6k12048




          26.6k12048








          • 1




            $begingroup$
            i actually want to know about the degree..
            $endgroup$
            – Aryaman Bansal
            Jan 24 '15 at 14:33










          • $begingroup$
            @AryamanBansal. First order (just as abel answered).
            $endgroup$
            – Claude Leibovici
            Jan 24 '15 at 14:35










          • $begingroup$
            @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
            $endgroup$
            – abel
            Jan 24 '15 at 14:38






          • 2




            $begingroup$
            @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
            $endgroup$
            – Singh
            Dec 22 '15 at 3:01






          • 1




            $begingroup$
            @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
            $endgroup$
            – Singh
            Dec 22 '15 at 3:42
















          • 1




            $begingroup$
            i actually want to know about the degree..
            $endgroup$
            – Aryaman Bansal
            Jan 24 '15 at 14:33










          • $begingroup$
            @AryamanBansal. First order (just as abel answered).
            $endgroup$
            – Claude Leibovici
            Jan 24 '15 at 14:35










          • $begingroup$
            @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
            $endgroup$
            – abel
            Jan 24 '15 at 14:38






          • 2




            $begingroup$
            @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
            $endgroup$
            – Singh
            Dec 22 '15 at 3:01






          • 1




            $begingroup$
            @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
            $endgroup$
            – Singh
            Dec 22 '15 at 3:42










          1




          1




          $begingroup$
          i actually want to know about the degree..
          $endgroup$
          – Aryaman Bansal
          Jan 24 '15 at 14:33




          $begingroup$
          i actually want to know about the degree..
          $endgroup$
          – Aryaman Bansal
          Jan 24 '15 at 14:33












          $begingroup$
          @AryamanBansal. First order (just as abel answered).
          $endgroup$
          – Claude Leibovici
          Jan 24 '15 at 14:35




          $begingroup$
          @AryamanBansal. First order (just as abel answered).
          $endgroup$
          – Claude Leibovici
          Jan 24 '15 at 14:35












          $begingroup$
          @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
          $endgroup$
          – abel
          Jan 24 '15 at 14:38




          $begingroup$
          @AryamanBansal, even if you are talking about degree, what is the degree of $sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree.
          $endgroup$
          – abel
          Jan 24 '15 at 14:38




          2




          2




          $begingroup$
          @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
          $endgroup$
          – Singh
          Dec 22 '15 at 3:01




          $begingroup$
          @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see.
          $endgroup$
          – Singh
          Dec 22 '15 at 3:01




          1




          1




          $begingroup$
          @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
          $endgroup$
          – Singh
          Dec 22 '15 at 3:42






          $begingroup$
          @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf)
          $endgroup$
          – Singh
          Dec 22 '15 at 3:42




















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