Nonhomogeneous Second oder ODE
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I've recently started to learn ode, consider this equation $t''+at'-t=f$, where $a$ is not a constant and $f$ is a desired function.
$$y^{2} + ay-1= 0 quadLeftrightarrowquad y_{pm} = - frac{a}{2} pm sqrt{left(frac{a}{2}right)^{!2} +1}, $$
so for simplicity let's call it $y_{+}=A$ and $y_{-}=B $ as fundamental solution.
if I'm not mistaken this is the general solution:
$$t= frac{fA}{-BA'+AB'}-frac{fB}{-BA'+AB'}+C_{1}A+C_{2}B.$$
Now I want to know if this can be an explicit solution or not? And how can I show existence and uniqueness of this solution?
ordinary-differential-equations
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add a comment |
$begingroup$
I've recently started to learn ode, consider this equation $t''+at'-t=f$, where $a$ is not a constant and $f$ is a desired function.
$$y^{2} + ay-1= 0 quadLeftrightarrowquad y_{pm} = - frac{a}{2} pm sqrt{left(frac{a}{2}right)^{!2} +1}, $$
so for simplicity let's call it $y_{+}=A$ and $y_{-}=B $ as fundamental solution.
if I'm not mistaken this is the general solution:
$$t= frac{fA}{-BA'+AB'}-frac{fB}{-BA'+AB'}+C_{1}A+C_{2}B.$$
Now I want to know if this can be an explicit solution or not? And how can I show existence and uniqueness of this solution?
ordinary-differential-equations
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The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
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– Yves Daoust
Jan 22 at 16:50
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@YvesDaoust So, what is the right solution for this equation? How can I solve it?
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– Angela
Jan 22 at 17:04
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In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
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– Yves Daoust
Jan 22 at 17:21
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@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
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– Angela
Jan 22 at 17:25
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No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53
add a comment |
$begingroup$
I've recently started to learn ode, consider this equation $t''+at'-t=f$, where $a$ is not a constant and $f$ is a desired function.
$$y^{2} + ay-1= 0 quadLeftrightarrowquad y_{pm} = - frac{a}{2} pm sqrt{left(frac{a}{2}right)^{!2} +1}, $$
so for simplicity let's call it $y_{+}=A$ and $y_{-}=B $ as fundamental solution.
if I'm not mistaken this is the general solution:
$$t= frac{fA}{-BA'+AB'}-frac{fB}{-BA'+AB'}+C_{1}A+C_{2}B.$$
Now I want to know if this can be an explicit solution or not? And how can I show existence and uniqueness of this solution?
ordinary-differential-equations
$endgroup$
I've recently started to learn ode, consider this equation $t''+at'-t=f$, where $a$ is not a constant and $f$ is a desired function.
$$y^{2} + ay-1= 0 quadLeftrightarrowquad y_{pm} = - frac{a}{2} pm sqrt{left(frac{a}{2}right)^{!2} +1}, $$
so for simplicity let's call it $y_{+}=A$ and $y_{-}=B $ as fundamental solution.
if I'm not mistaken this is the general solution:
$$t= frac{fA}{-BA'+AB'}-frac{fB}{-BA'+AB'}+C_{1}A+C_{2}B.$$
Now I want to know if this can be an explicit solution or not? And how can I show existence and uniqueness of this solution?
ordinary-differential-equations
ordinary-differential-equations
edited Jan 22 at 15:56
Adrian Keister
5,27371933
5,27371933
asked Jan 22 at 15:49
AngelaAngela
61
61
$begingroup$
The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
$endgroup$
– Yves Daoust
Jan 22 at 16:50
$begingroup$
@YvesDaoust So, what is the right solution for this equation? How can I solve it?
$endgroup$
– Angela
Jan 22 at 17:04
$begingroup$
In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
$endgroup$
– Yves Daoust
Jan 22 at 17:21
$begingroup$
@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
$endgroup$
– Angela
Jan 22 at 17:25
$begingroup$
No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53
add a comment |
$begingroup$
The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
$endgroup$
– Yves Daoust
Jan 22 at 16:50
$begingroup$
@YvesDaoust So, what is the right solution for this equation? How can I solve it?
$endgroup$
– Angela
Jan 22 at 17:04
$begingroup$
In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
$endgroup$
– Yves Daoust
Jan 22 at 17:21
$begingroup$
@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
$endgroup$
– Angela
Jan 22 at 17:25
$begingroup$
No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53
$begingroup$
The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
$endgroup$
– Yves Daoust
Jan 22 at 16:50
$begingroup$
The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
$endgroup$
– Yves Daoust
Jan 22 at 16:50
$begingroup$
@YvesDaoust So, what is the right solution for this equation? How can I solve it?
$endgroup$
– Angela
Jan 22 at 17:04
$begingroup$
@YvesDaoust So, what is the right solution for this equation? How can I solve it?
$endgroup$
– Angela
Jan 22 at 17:04
$begingroup$
In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
$endgroup$
– Yves Daoust
Jan 22 at 17:21
$begingroup$
In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
$endgroup$
– Yves Daoust
Jan 22 at 17:21
$begingroup$
@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
$endgroup$
– Angela
Jan 22 at 17:25
$begingroup$
@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
$endgroup$
– Angela
Jan 22 at 17:25
$begingroup$
No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53
$begingroup$
No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53
add a comment |
1 Answer
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I would try a series solution - there isn't enough information on a, besides it being an arbitrary function.
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1 Answer
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1 Answer
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I would try a series solution - there isn't enough information on a, besides it being an arbitrary function.
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add a comment |
$begingroup$
I would try a series solution - there isn't enough information on a, besides it being an arbitrary function.
$endgroup$
add a comment |
$begingroup$
I would try a series solution - there isn't enough information on a, besides it being an arbitrary function.
$endgroup$
I would try a series solution - there isn't enough information on a, besides it being an arbitrary function.
answered Jan 23 at 2:53


FizzerFizzer
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$begingroup$
The characteristic polynomial method only works for constant coefficients. (And the rest of the development is also wrong).
$endgroup$
– Yves Daoust
Jan 22 at 16:50
$begingroup$
@YvesDaoust So, what is the right solution for this equation? How can I solve it?
$endgroup$
– Angela
Jan 22 at 17:04
$begingroup$
In the general case, solving such a second order ODE equation is painful. As it is linear, you first solve the homogeneous part. Then you can try variation of constants to obtain a particular solution.
$endgroup$
– Yves Daoust
Jan 22 at 17:21
$begingroup$
@YvesDaoust and what is fundamental solution for homogeneous equation? It's $e^{y_{/pm}}t$
$endgroup$
– Angela
Jan 22 at 17:25
$begingroup$
No. Lookup "second order linear ODE with variable coefficients".
$endgroup$
– Yves Daoust
Jan 22 at 17:53