How do I convert this second order differential equation to two first order differential equations?
$begingroup$
The equation is $$m{ddot x} + kx + gsinθ = 0.$$
I know I have to convert it to the form ${dot y}_1 = y_2$ and ${dot y}_2 = text{something}$.
However I am very inexperienced and very confused on how to find $y_1$ and $y_2$ from this initial equation.
calculus linear-algebra integration derivatives
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
$endgroup$
add a comment |
$begingroup$
The equation is $$m{ddot x} + kx + gsinθ = 0.$$
I know I have to convert it to the form ${dot y}_1 = y_2$ and ${dot y}_2 = text{something}$.
However I am very inexperienced and very confused on how to find $y_1$ and $y_2$ from this initial equation.
calculus linear-algebra integration derivatives
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
$endgroup$
add a comment |
$begingroup$
The equation is $$m{ddot x} + kx + gsinθ = 0.$$
I know I have to convert it to the form ${dot y}_1 = y_2$ and ${dot y}_2 = text{something}$.
However I am very inexperienced and very confused on how to find $y_1$ and $y_2$ from this initial equation.
calculus linear-algebra integration derivatives
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
$endgroup$
The equation is $$m{ddot x} + kx + gsinθ = 0.$$
I know I have to convert it to the form ${dot y}_1 = y_2$ and ${dot y}_2 = text{something}$.
However I am very inexperienced and very confused on how to find $y_1$ and $y_2$ from this initial equation.
calculus linear-algebra integration derivatives
calculus linear-algebra integration derivatives
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
edited Dec 13 '18 at 22:45
Tianlalu
3,08421138
3,08421138
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
asked Dec 13 '18 at 22:35
yeetuscleetusyeetuscleetus
41
41
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's pose $dot{x} = v$ (i.e. velocity). Then:
$$begin{cases}
m dot{v} + kx + gsintheta = 0\
dot{x} = v
end{cases} Rightarrow begin{cases}
dot{v} = -frac{k}{m}x - frac{g}{m}sintheta\
dot{x} = v.
end{cases}$$
Maybe, further calculation can be done for $theta$. But if you don't specify its meaning, then these just are meaningless suppositions.
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's pose $dot{x} = v$ (i.e. velocity). Then:
$$begin{cases}
m dot{v} + kx + gsintheta = 0\
dot{x} = v
end{cases} Rightarrow begin{cases}
dot{v} = -frac{k}{m}x - frac{g}{m}sintheta\
dot{x} = v.
end{cases}$$
Maybe, further calculation can be done for $theta$. But if you don't specify its meaning, then these just are meaningless suppositions.
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
add a comment |
$begingroup$
Let's pose $dot{x} = v$ (i.e. velocity). Then:
$$begin{cases}
m dot{v} + kx + gsintheta = 0\
dot{x} = v
end{cases} Rightarrow begin{cases}
dot{v} = -frac{k}{m}x - frac{g}{m}sintheta\
dot{x} = v.
end{cases}$$
Maybe, further calculation can be done for $theta$. But if you don't specify its meaning, then these just are meaningless suppositions.
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
$endgroup$
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
add a comment |
$begingroup$
Let's pose $dot{x} = v$ (i.e. velocity). Then:
$$begin{cases}
m dot{v} + kx + gsintheta = 0\
dot{x} = v
end{cases} Rightarrow begin{cases}
dot{v} = -frac{k}{m}x - frac{g}{m}sintheta\
dot{x} = v.
end{cases}$$
Maybe, further calculation can be done for $theta$. But if you don't specify its meaning, then these just are meaningless suppositions.
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
$endgroup$
Let's pose $dot{x} = v$ (i.e. velocity). Then:
$$begin{cases}
m dot{v} + kx + gsintheta = 0\
dot{x} = v
end{cases} Rightarrow begin{cases}
dot{v} = -frac{k}{m}x - frac{g}{m}sintheta\
dot{x} = v.
end{cases}$$
Maybe, further calculation can be done for $theta$. But if you don't specify its meaning, then these just are meaningless suppositions.
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
edited Dec 13 '18 at 23:04
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
answered Dec 13 '18 at 22:38
the_candymanthe_candyman
8,97832145
8,97832145
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
add a comment |
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
$begingroup$
θ is an arbitrary, constant angle, that represents the angle above horizontal for a slanted plane. I am not experienced enough to know if that will affect the solution to the linear system.
$endgroup$
– yeetuscleetus
Dec 13 '18 at 22:46
1
1
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
@erics: The value of the parameter $theta$ certainly affects the solution set: If $sin theta = 0$, the system is homogeneous, and in particular $x(t) = 0$ is a solution (this makes good intuitive sense in terms of the physical system, too). If not, the system is not homogeneous and $x(t) = 0$ is not a solution.
$endgroup$
– Travis
Dec 13 '18 at 22:54
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
is there any relationship between $theta$ and $x$? I mean, can I write $theta$ as a function of $x$ (or vice versa)? If not, my answer is complete.
$endgroup$
– the_candyman
Dec 13 '18 at 23:02
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
$begingroup$
@the_candyman No, there is no relationship. Your answer is correct, thank you!
$endgroup$
– yeetuscleetus
Dec 14 '18 at 0:20
add a comment |
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