How to interpret the differential equation for the sine function?












0














I have read that the sine function can be defined as y'' = -y , i.e., the rate of rate of change of the function at any point is the negative of the value of the function at that point.



I am not able to visualize this. If the rate of rate of change is negative of the value at any point, won't the function immediately bounce back to zero at any point?






differential-equations







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  • Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
    – LutzL
    Jan 4 at 16:49








  • 2




    That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
    – John Douma
    Jan 4 at 16:51
















0














I have read that the sine function can be defined as y'' = -y , i.e., the rate of rate of change of the function at any point is the negative of the value of the function at that point.



I am not able to visualize this. If the rate of rate of change is negative of the value at any point, won't the function immediately bounce back to zero at any point?






differential-equations







share|cite|improve this question







New contributor




Curious123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
    – LutzL
    Jan 4 at 16:49








  • 2




    That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
    – John Douma
    Jan 4 at 16:51














0












0








0







I have read that the sine function can be defined as y'' = -y , i.e., the rate of rate of change of the function at any point is the negative of the value of the function at that point.



I am not able to visualize this. If the rate of rate of change is negative of the value at any point, won't the function immediately bounce back to zero at any point?






differential-equations







share|cite|improve this question







New contributor




Curious123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have read that the sine function can be defined as y'' = -y , i.e., the rate of rate of change of the function at any point is the negative of the value of the function at that point.



I am not able to visualize this. If the rate of rate of change is negative of the value at any point, won't the function immediately bounce back to zero at any point?






differential-equations




differential-equations






share|cite|improve this question







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Curious123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







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asked Jan 4 at 16:45









Curious123Curious123

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  • Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
    – LutzL
    Jan 4 at 16:49








  • 2




    That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
    – John Douma
    Jan 4 at 16:51


















  • Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
    – LutzL
    Jan 4 at 16:49








  • 2




    That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
    – John Douma
    Jan 4 at 16:51
















Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
– LutzL
Jan 4 at 16:49






Earth's orbit follows the law $ddot{frak x}=-cdfrac{frak x}{|{frak x}|^3}$ which is similar, but did the Earth fall into the sun in the eons of its existence?
– LutzL
Jan 4 at 16:49






2




2




That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
– John Douma
Jan 4 at 16:51




That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
– John Douma
Jan 4 at 16:51










2 Answers
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1














The equation says that the second derivative of $y$ equals the negative of $y$ and this is intuitive if we remember what the second derivative tells us.



When $ygt 0$ the sine is concave down and when $ylt 0$ the sine is concave up. The inflection points are where $y=0$.






share|cite|improve this answer





























    0














    Multiply with $2y'$ and integrate to find that
    $$
    frac{d}{dt}(y'^2+y^2)=2y'(y''+y)=0
    $$
    so that the radius of the point $(y',y)$ is a constant $R$. As such any such point can be parametrized by trigonometric functions whose angle varies with time, $y'=Rcosphi$, $y=Rsinphi$. Now compare the derivative of the second equation with the first equation to find that
    $$
    Rcosphi=y'=Rcosphicdotphi'.
    $$
    This means that the rate of change of the angle is constant $1$, $phi(t)=t+phi_0$, $$y=Rsin(phi_0+t).$$




    Or look at the infinitesimal picture,
    $$
    y(t+dt)-2y(t)+y(t-dt)=-y(t),dt^2+O(dt^4)
    $$
    The characteristic equation is $0=q^2-(2-dt^2)q+1=(q-1+dt^2/2)^2+dt^2-dt^4/4$ so again modulo $O(dt^4)$ we get $q=1-dt^2/2pm i,dt=cos(dt)+isin(dt)+O(dt^3)$ and the basis solutions by the Euler-Moivre formula are
    $$
    y_{1,2}(t_0+n,dt)=q^n=cos(n,dt)pm isin(n,dt)+O(dt^2)=cos(t-t_0)pm isin(t-t_0)+O(dt^2)
    $$
    and in the limit $dtto 0$, so that also on this way one gets the general solution $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      active

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      1














      The equation says that the second derivative of $y$ equals the negative of $y$ and this is intuitive if we remember what the second derivative tells us.



      When $ygt 0$ the sine is concave down and when $ylt 0$ the sine is concave up. The inflection points are where $y=0$.






      share|cite|improve this answer


























        1














        The equation says that the second derivative of $y$ equals the negative of $y$ and this is intuitive if we remember what the second derivative tells us.



        When $ygt 0$ the sine is concave down and when $ylt 0$ the sine is concave up. The inflection points are where $y=0$.






        share|cite|improve this answer
























          1












          1








          1






          The equation says that the second derivative of $y$ equals the negative of $y$ and this is intuitive if we remember what the second derivative tells us.



          When $ygt 0$ the sine is concave down and when $ylt 0$ the sine is concave up. The inflection points are where $y=0$.






          share|cite|improve this answer












          The equation says that the second derivative of $y$ equals the negative of $y$ and this is intuitive if we remember what the second derivative tells us.



          When $ygt 0$ the sine is concave down and when $ylt 0$ the sine is concave up. The inflection points are where $y=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 18:21









          John DoumaJohn Douma

          5,41211319




          5,41211319























              0














              Multiply with $2y'$ and integrate to find that
              $$
              frac{d}{dt}(y'^2+y^2)=2y'(y''+y)=0
              $$
              so that the radius of the point $(y',y)$ is a constant $R$. As such any such point can be parametrized by trigonometric functions whose angle varies with time, $y'=Rcosphi$, $y=Rsinphi$. Now compare the derivative of the second equation with the first equation to find that
              $$
              Rcosphi=y'=Rcosphicdotphi'.
              $$
              This means that the rate of change of the angle is constant $1$, $phi(t)=t+phi_0$, $$y=Rsin(phi_0+t).$$




              Or look at the infinitesimal picture,
              $$
              y(t+dt)-2y(t)+y(t-dt)=-y(t),dt^2+O(dt^4)
              $$
              The characteristic equation is $0=q^2-(2-dt^2)q+1=(q-1+dt^2/2)^2+dt^2-dt^4/4$ so again modulo $O(dt^4)$ we get $q=1-dt^2/2pm i,dt=cos(dt)+isin(dt)+O(dt^3)$ and the basis solutions by the Euler-Moivre formula are
              $$
              y_{1,2}(t_0+n,dt)=q^n=cos(n,dt)pm isin(n,dt)+O(dt^2)=cos(t-t_0)pm isin(t-t_0)+O(dt^2)
              $$
              and in the limit $dtto 0$, so that also on this way one gets the general solution $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$






              share|cite|improve this answer


























                0














                Multiply with $2y'$ and integrate to find that
                $$
                frac{d}{dt}(y'^2+y^2)=2y'(y''+y)=0
                $$
                so that the radius of the point $(y',y)$ is a constant $R$. As such any such point can be parametrized by trigonometric functions whose angle varies with time, $y'=Rcosphi$, $y=Rsinphi$. Now compare the derivative of the second equation with the first equation to find that
                $$
                Rcosphi=y'=Rcosphicdotphi'.
                $$
                This means that the rate of change of the angle is constant $1$, $phi(t)=t+phi_0$, $$y=Rsin(phi_0+t).$$




                Or look at the infinitesimal picture,
                $$
                y(t+dt)-2y(t)+y(t-dt)=-y(t),dt^2+O(dt^4)
                $$
                The characteristic equation is $0=q^2-(2-dt^2)q+1=(q-1+dt^2/2)^2+dt^2-dt^4/4$ so again modulo $O(dt^4)$ we get $q=1-dt^2/2pm i,dt=cos(dt)+isin(dt)+O(dt^3)$ and the basis solutions by the Euler-Moivre formula are
                $$
                y_{1,2}(t_0+n,dt)=q^n=cos(n,dt)pm isin(n,dt)+O(dt^2)=cos(t-t_0)pm isin(t-t_0)+O(dt^2)
                $$
                and in the limit $dtto 0$, so that also on this way one gets the general solution $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Multiply with $2y'$ and integrate to find that
                  $$
                  frac{d}{dt}(y'^2+y^2)=2y'(y''+y)=0
                  $$
                  so that the radius of the point $(y',y)$ is a constant $R$. As such any such point can be parametrized by trigonometric functions whose angle varies with time, $y'=Rcosphi$, $y=Rsinphi$. Now compare the derivative of the second equation with the first equation to find that
                  $$
                  Rcosphi=y'=Rcosphicdotphi'.
                  $$
                  This means that the rate of change of the angle is constant $1$, $phi(t)=t+phi_0$, $$y=Rsin(phi_0+t).$$




                  Or look at the infinitesimal picture,
                  $$
                  y(t+dt)-2y(t)+y(t-dt)=-y(t),dt^2+O(dt^4)
                  $$
                  The characteristic equation is $0=q^2-(2-dt^2)q+1=(q-1+dt^2/2)^2+dt^2-dt^4/4$ so again modulo $O(dt^4)$ we get $q=1-dt^2/2pm i,dt=cos(dt)+isin(dt)+O(dt^3)$ and the basis solutions by the Euler-Moivre formula are
                  $$
                  y_{1,2}(t_0+n,dt)=q^n=cos(n,dt)pm isin(n,dt)+O(dt^2)=cos(t-t_0)pm isin(t-t_0)+O(dt^2)
                  $$
                  and in the limit $dtto 0$, so that also on this way one gets the general solution $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$






                  share|cite|improve this answer












                  Multiply with $2y'$ and integrate to find that
                  $$
                  frac{d}{dt}(y'^2+y^2)=2y'(y''+y)=0
                  $$
                  so that the radius of the point $(y',y)$ is a constant $R$. As such any such point can be parametrized by trigonometric functions whose angle varies with time, $y'=Rcosphi$, $y=Rsinphi$. Now compare the derivative of the second equation with the first equation to find that
                  $$
                  Rcosphi=y'=Rcosphicdotphi'.
                  $$
                  This means that the rate of change of the angle is constant $1$, $phi(t)=t+phi_0$, $$y=Rsin(phi_0+t).$$




                  Or look at the infinitesimal picture,
                  $$
                  y(t+dt)-2y(t)+y(t-dt)=-y(t),dt^2+O(dt^4)
                  $$
                  The characteristic equation is $0=q^2-(2-dt^2)q+1=(q-1+dt^2/2)^2+dt^2-dt^4/4$ so again modulo $O(dt^4)$ we get $q=1-dt^2/2pm i,dt=cos(dt)+isin(dt)+O(dt^3)$ and the basis solutions by the Euler-Moivre formula are
                  $$
                  y_{1,2}(t_0+n,dt)=q^n=cos(n,dt)pm isin(n,dt)+O(dt^2)=cos(t-t_0)pm isin(t-t_0)+O(dt^2)
                  $$
                  and in the limit $dtto 0$, so that also on this way one gets the general solution $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 20:27









                  LutzLLutzL

                  56.5k42054




                  56.5k42054






















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