Solution to a second-order nonlinear ordinary differential equation
$begingroup$
I am given three functions $z'(t), y'(t), x'(t)$ and I have reduced these functions to the equation:
begin{equation}
z''=x_{0}beta z'e^{-frac{beta}{gamma}z} - gamma z
end{equation}
where $beta, gamma ,x_{0}$ are constants.
Next we introduce a new function:
begin{equation}
u(t)=e^{-frac{beta}{gamma}z}
end{equation}
Substitution yields
begin{equation}
ufrac{d^{2}u}{dt^{2}}- bigg(frac{du}{dt}bigg)^{2}+(gamma - x_{0}beta u)ufrac{du}{dt} = 0
end{equation}
Next a new function is introduced:
begin{equation}
phi = frac{dt}{du}
end{equation}
Now the equation can be rearranged to:
begin{equation}
frac{dphi}{du}+frac{1}{u} phi = (gamma-x_{0}beta u )phi^{2}
end{equation}
The solution to this equation is:
begin{equation}
phi = frac{1}{u(C_{1}-gamma ln u +x_{0}beta u)}
end{equation}
So what would be the solution to the original equation in terms of $u$?
real-analysis calculus ordinary-differential-equations
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
$endgroup$
add a comment |
$begingroup$
I am given three functions $z'(t), y'(t), x'(t)$ and I have reduced these functions to the equation:
begin{equation}
z''=x_{0}beta z'e^{-frac{beta}{gamma}z} - gamma z
end{equation}
where $beta, gamma ,x_{0}$ are constants.
Next we introduce a new function:
begin{equation}
u(t)=e^{-frac{beta}{gamma}z}
end{equation}
Substitution yields
begin{equation}
ufrac{d^{2}u}{dt^{2}}- bigg(frac{du}{dt}bigg)^{2}+(gamma - x_{0}beta u)ufrac{du}{dt} = 0
end{equation}
Next a new function is introduced:
begin{equation}
phi = frac{dt}{du}
end{equation}
Now the equation can be rearranged to:
begin{equation}
frac{dphi}{du}+frac{1}{u} phi = (gamma-x_{0}beta u )phi^{2}
end{equation}
The solution to this equation is:
begin{equation}
phi = frac{1}{u(C_{1}-gamma ln u +x_{0}beta u)}
end{equation}
So what would be the solution to the original equation in terms of $u$?
real-analysis calculus ordinary-differential-equations
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
$endgroup$
$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31
add a comment |
$begingroup$
I am given three functions $z'(t), y'(t), x'(t)$ and I have reduced these functions to the equation:
begin{equation}
z''=x_{0}beta z'e^{-frac{beta}{gamma}z} - gamma z
end{equation}
where $beta, gamma ,x_{0}$ are constants.
Next we introduce a new function:
begin{equation}
u(t)=e^{-frac{beta}{gamma}z}
end{equation}
Substitution yields
begin{equation}
ufrac{d^{2}u}{dt^{2}}- bigg(frac{du}{dt}bigg)^{2}+(gamma - x_{0}beta u)ufrac{du}{dt} = 0
end{equation}
Next a new function is introduced:
begin{equation}
phi = frac{dt}{du}
end{equation}
Now the equation can be rearranged to:
begin{equation}
frac{dphi}{du}+frac{1}{u} phi = (gamma-x_{0}beta u )phi^{2}
end{equation}
The solution to this equation is:
begin{equation}
phi = frac{1}{u(C_{1}-gamma ln u +x_{0}beta u)}
end{equation}
So what would be the solution to the original equation in terms of $u$?
real-analysis calculus ordinary-differential-equations
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
$endgroup$
I am given three functions $z'(t), y'(t), x'(t)$ and I have reduced these functions to the equation:
begin{equation}
z''=x_{0}beta z'e^{-frac{beta}{gamma}z} - gamma z
end{equation}
where $beta, gamma ,x_{0}$ are constants.
Next we introduce a new function:
begin{equation}
u(t)=e^{-frac{beta}{gamma}z}
end{equation}
Substitution yields
begin{equation}
ufrac{d^{2}u}{dt^{2}}- bigg(frac{du}{dt}bigg)^{2}+(gamma - x_{0}beta u)ufrac{du}{dt} = 0
end{equation}
Next a new function is introduced:
begin{equation}
phi = frac{dt}{du}
end{equation}
Now the equation can be rearranged to:
begin{equation}
frac{dphi}{du}+frac{1}{u} phi = (gamma-x_{0}beta u )phi^{2}
end{equation}
The solution to this equation is:
begin{equation}
phi = frac{1}{u(C_{1}-gamma ln u +x_{0}beta u)}
end{equation}
So what would be the solution to the original equation in terms of $u$?
real-analysis calculus ordinary-differential-equations
real-analysis calculus ordinary-differential-equations
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
asked Jan 15 at 17:10
wittgensteinsrulerwittgensteinsruler
243
243
$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31
add a comment |
$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31
$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31
$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The last equation od
$$
frac{dt}{du}=frac{1}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
Integrating
$$
t=intfrac{du}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
I do not think that there is a closed formula for the integral.
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
The last equation od
$$
frac{dt}{du}=frac{1}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
Integrating
$$
t=intfrac{du}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
I do not think that there is a closed formula for the integral.
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
add a comment |
$begingroup$
The last equation od
$$
frac{dt}{du}=frac{1}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
Integrating
$$
t=intfrac{du}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
I do not think that there is a closed formula for the integral.
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
add a comment |
$begingroup$
The last equation od
$$
frac{dt}{du}=frac{1}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
Integrating
$$
t=intfrac{du}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
I do not think that there is a closed formula for the integral.
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
The last equation od
$$
frac{dt}{du}=frac{1}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
Integrating
$$
t=intfrac{du}{u(C_{1}-gamma ln u +x_{0}beta, u)}.
$$
I do not think that there is a closed formula for the integral.
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 15 at 22:52
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
add a comment |
add a comment |
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$begingroup$
The ordinary differential equation (ODE) that you write down for $u$ is equivalent to $z'' = x_0 beta z' e^{-frac{beta}{gamma} z} - gamma z'$. So there is either a mistake in the original ODE for $z$ or in the ODE for $u$. Please check.
$endgroup$
– Christoph
Jan 16 at 2:31