How to interpret the differential equation for the sine function?
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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?
ordinary-differential-equations
asked Jan 4 at 16:45
Curious123Curious123
11
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add a comment |
$begingroup$
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?
ordinary-differential-equations
asked Jan 4 at 16:45
Curious123Curious123
11
$endgroup$
$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
2
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51
add a comment |
$begingroup$
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?
ordinary-differential-equations
asked Jan 4 at 16:45
Curious123Curious123
11
$endgroup$
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?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 4 at 16:45
Curious123Curious123
11
asked Jan 4 at 16:45
Curious123Curious123
11
asked Jan 4 at 16:45
Curious123Curious123
11
asked Jan 4 at 16:45
Curious123Curious123
11
asked Jan 4 at 16:45
Curious123Curious123
11
11
$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
2
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51
add a comment |
$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
2
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51
$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
2
2
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51
add a comment |
2 Answers
2
active
oldest
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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 $y=0$.
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
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add a comment |
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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 $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $y=0$.
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
$endgroup$
add a comment |
$begingroup$
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 $y=0$.
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
$endgroup$
add a comment |
$begingroup$
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 $y=0$.
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
$endgroup$
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 $y=0$.
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
answered Jan 4 at 18:21
John DoumaJohn Douma
5,51211319
5,51211319
add a comment |
add a comment |
$begingroup$
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 $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
$endgroup$
add a comment |
$begingroup$
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 $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
$endgroup$
add a comment |
$begingroup$
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 $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
$endgroup$
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 $$y(t)=y'(t_0)sin(t-t_0)+y(t_0)cos(t-t_0).$$
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
answered Jan 4 at 20:27
LutzLLutzL
57.8k42054
57.8k42054
add a comment |
add a comment |
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$begingroup$
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?
$endgroup$
– LutzL
Jan 4 at 16:49
2
$begingroup$
That differential equation does not say that the rate of change is the negative of $y$. Notice the double prime.
$endgroup$
– John Douma
Jan 4 at 16:51