Difficult Second Order Nonlinear Differential Equation
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I've been attempting to create a model for a particular physical phenomena. I've reached the stage where I need to find the solution to a differential equation in order to continue. I've attempted to solve it on my own through a substitution but was unable to get anywhere with it. I am aware that the differential equation may not solvable so I will provide a few different forms. If any of these are solvable please let me know.
a) $y'' + (y')^2 = sin(y) $
b) $y'' + (y')^2 = y $
Thank you for your time and help.
ordinary-differential-equations
asked Dec 23 '18 at 18:52
StavrosStavros
62
$endgroup$
add a comment |
$begingroup$
I've been attempting to create a model for a particular physical phenomena. I've reached the stage where I need to find the solution to a differential equation in order to continue. I've attempted to solve it on my own through a substitution but was unable to get anywhere with it. I am aware that the differential equation may not solvable so I will provide a few different forms. If any of these are solvable please let me know.
a) $y'' + (y')^2 = sin(y) $
b) $y'' + (y')^2 = y $
Thank you for your time and help.
ordinary-differential-equations
asked Dec 23 '18 at 18:52
StavrosStavros
62
$endgroup$
$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
1
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08
add a comment |
$begingroup$
I've been attempting to create a model for a particular physical phenomena. I've reached the stage where I need to find the solution to a differential equation in order to continue. I've attempted to solve it on my own through a substitution but was unable to get anywhere with it. I am aware that the differential equation may not solvable so I will provide a few different forms. If any of these are solvable please let me know.
a) $y'' + (y')^2 = sin(y) $
b) $y'' + (y')^2 = y $
Thank you for your time and help.
ordinary-differential-equations
asked Dec 23 '18 at 18:52
StavrosStavros
62
$endgroup$
I've been attempting to create a model for a particular physical phenomena. I've reached the stage where I need to find the solution to a differential equation in order to continue. I've attempted to solve it on my own through a substitution but was unable to get anywhere with it. I am aware that the differential equation may not solvable so I will provide a few different forms. If any of these are solvable please let me know.
a) $y'' + (y')^2 = sin(y) $
b) $y'' + (y')^2 = y $
Thank you for your time and help.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 23 '18 at 18:52
StavrosStavros
62
asked Dec 23 '18 at 18:52
StavrosStavros
62
asked Dec 23 '18 at 18:52
StavrosStavros
62
asked Dec 23 '18 at 18:52
StavrosStavros
62
asked Dec 23 '18 at 18:52
StavrosStavros
62
62
$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
1
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08
add a comment |
$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
1
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08
$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
1
1
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $u = (y')^2$. You can transform as follows:
$$ y'' = frac{frac{d}{dx}(y'^2)}{2y'} = frac12frac{dfrac{du}{dx}}{dfrac{dy}{dx}} = frac12frac{du}{dy} $$
Then
$$ frac{du}{dy} + 2u = 2g(y) $$
where $g(y)$ is any function on your RHS. This is a standard first-order equation in $u(y)$.
Once you have $u(y)$, it remains to solve
$$ frac{dy}{dx} = pmsqrt{u(y)} $$
which is a separable equation
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
$endgroup$
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
add a comment |
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1 Answer
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$begingroup$
Let $u = (y')^2$. You can transform as follows:
$$ y'' = frac{frac{d}{dx}(y'^2)}{2y'} = frac12frac{dfrac{du}{dx}}{dfrac{dy}{dx}} = frac12frac{du}{dy} $$
Then
$$ frac{du}{dy} + 2u = 2g(y) $$
where $g(y)$ is any function on your RHS. This is a standard first-order equation in $u(y)$.
Once you have $u(y)$, it remains to solve
$$ frac{dy}{dx} = pmsqrt{u(y)} $$
which is a separable equation
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
$endgroup$
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
add a comment |
$begingroup$
Let $u = (y')^2$. You can transform as follows:
$$ y'' = frac{frac{d}{dx}(y'^2)}{2y'} = frac12frac{dfrac{du}{dx}}{dfrac{dy}{dx}} = frac12frac{du}{dy} $$
Then
$$ frac{du}{dy} + 2u = 2g(y) $$
where $g(y)$ is any function on your RHS. This is a standard first-order equation in $u(y)$.
Once you have $u(y)$, it remains to solve
$$ frac{dy}{dx} = pmsqrt{u(y)} $$
which is a separable equation
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
$endgroup$
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
add a comment |
$begingroup$
Let $u = (y')^2$. You can transform as follows:
$$ y'' = frac{frac{d}{dx}(y'^2)}{2y'} = frac12frac{dfrac{du}{dx}}{dfrac{dy}{dx}} = frac12frac{du}{dy} $$
Then
$$ frac{du}{dy} + 2u = 2g(y) $$
where $g(y)$ is any function on your RHS. This is a standard first-order equation in $u(y)$.
Once you have $u(y)$, it remains to solve
$$ frac{dy}{dx} = pmsqrt{u(y)} $$
which is a separable equation
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
$endgroup$
Let $u = (y')^2$. You can transform as follows:
$$ y'' = frac{frac{d}{dx}(y'^2)}{2y'} = frac12frac{dfrac{du}{dx}}{dfrac{dy}{dx}} = frac12frac{du}{dy} $$
Then
$$ frac{du}{dy} + 2u = 2g(y) $$
where $g(y)$ is any function on your RHS. This is a standard first-order equation in $u(y)$.
Once you have $u(y)$, it remains to solve
$$ frac{dy}{dx} = pmsqrt{u(y)} $$
which is a separable equation
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
answered Dec 24 '18 at 4:00
DylanDylan
13.6k31027
13.6k31027
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
add a comment |
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It would appear that when $g(y) = sin(y) $ or $ g(y) = y$, it is not possible to integrate $u(y)$.
$endgroup$
– Stavros
Dec 24 '18 at 9:56
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
$begingroup$
It usually isn't possible to represent the solution in closed form. If you have some initial conditions, you can use numerical integration.
$endgroup$
– Dylan
Dec 24 '18 at 10:05
add a comment |
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$begingroup$
Do you have initial or boundary conditions? Are you considering numerical solutions or only analytical solutions?
$endgroup$
– rafa11111
Dec 23 '18 at 19:01
$begingroup$
I had an analytical solution in mind. However, if this is not possible a numerical solution would suffice.
$endgroup$
– Stavros
Dec 23 '18 at 19:05
1
$begingroup$
Note that $y''=dfrac{mathrm{d}}{mathrm{d}y}(y'^2/2)$ to reduce to a first-order linear equation in $y'^2$...
$endgroup$
– user10354138
Dec 23 '18 at 19:22
$begingroup$
If you just want the general behavior of the solution around a point $x_0$, one may try to do local analysis such as dominated balance to obtain an approximate solution.
$endgroup$
– Story123
Dec 23 '18 at 21:08