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From system of coupled ODEs to separable ODE
System of Nonlinear ODEs.Bounded solutions of a linear non-autonomous ODESolve linear system of ODEs using Laplace transformConverting an ODE into a system of first order ODEsPhase portrait of system of nonlinear ODEsFinding a second order ODE from a system of equationsODE system: change 't' to get a linear and autonomous systemHow to derive the higher order linear ODE from autonomous $1$st order ODE?Solving system of ODEsSolving an ODE system which depends on a periodic function
$begingroup$
How does one go from
beginalign
dotx&=y\
doty&=-x^3
endalign
to the following ODE?
$$fracdydx = -fracx^3y$$
ordinary-differential-equations
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
$endgroup$
add a comment |
$begingroup$
How does one go from
beginalign
dotx&=y\
doty&=-x^3
endalign
to the following ODE?
$$fracdydx = -fracx^3y$$
ordinary-differential-equations
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
$endgroup$
2
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago
add a comment |
$begingroup$
How does one go from
beginalign
dotx&=y\
doty&=-x^3
endalign
to the following ODE?
$$fracdydx = -fracx^3y$$
ordinary-differential-equations
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
$endgroup$
How does one go from
beginalign
dotx&=y\
doty&=-x^3
endalign
to the following ODE?
$$fracdydx = -fracx^3y$$
ordinary-differential-equations
ordinary-differential-equations
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
edited 8 hours ago
Rodrigo de Azevedo
13.4k42065
13.4k42065
asked 8 hours ago
ParsevalParseval
3,2061720
asked 8 hours ago
ParsevalParseval
3,2061720
asked 8 hours ago
ParsevalParseval
3,2061720
3,2061720
2
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago
add a comment |
2
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago
2
2
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Notice that:
$$dotx = y Rightarrow dx = ydt,$$
and
$$doty = -x^3 Rightarrow dy = -x^3 dt.$$
Therefore:
$$fracdydx = frac-x^3dtydt = -fracx^3y$$.
answered 8 hours ago
the_candymanthe_candyman
9,41732147
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracfracdydtfracdxdt=fracdydx$$
answered 8 hours ago
E.H.EE.H.E
18.2k12070
$endgroup$
add a comment |
$begingroup$
We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.
answered 8 hours ago
auscryptauscrypt
4,326212
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that:
$$dotx = y Rightarrow dx = ydt,$$
and
$$doty = -x^3 Rightarrow dy = -x^3 dt.$$
Therefore:
$$fracdydx = frac-x^3dtydt = -fracx^3y$$.
answered 8 hours ago
the_candymanthe_candyman
9,41732147
$endgroup$
add a comment |
$begingroup$
Notice that:
$$dotx = y Rightarrow dx = ydt,$$
and
$$doty = -x^3 Rightarrow dy = -x^3 dt.$$
Therefore:
$$fracdydx = frac-x^3dtydt = -fracx^3y$$.
answered 8 hours ago
the_candymanthe_candyman
9,41732147
$endgroup$
add a comment |
$begingroup$
Notice that:
$$dotx = y Rightarrow dx = ydt,$$
and
$$doty = -x^3 Rightarrow dy = -x^3 dt.$$
Therefore:
$$fracdydx = frac-x^3dtydt = -fracx^3y$$.
answered 8 hours ago
the_candymanthe_candyman
9,41732147
$endgroup$
Notice that:
$$dotx = y Rightarrow dx = ydt,$$
and
$$doty = -x^3 Rightarrow dy = -x^3 dt.$$
Therefore:
$$fracdydx = frac-x^3dtydt = -fracx^3y$$.
answered 8 hours ago
the_candymanthe_candyman
9,41732147
answered 8 hours ago
the_candymanthe_candyman
9,41732147
answered 8 hours ago
the_candymanthe_candyman
9,41732147
answered 8 hours ago
the_candymanthe_candyman
9,41732147
9,41732147
add a comment |
add a comment |
$begingroup$
Hint:
$$fracfracdydtfracdxdt=fracdydx$$
answered 8 hours ago
E.H.EE.H.E
18.2k12070
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracfracdydtfracdxdt=fracdydx$$
answered 8 hours ago
E.H.EE.H.E
18.2k12070
$endgroup$
add a comment |
$begingroup$
Hint:
$$fracfracdydtfracdxdt=fracdydx$$
answered 8 hours ago
E.H.EE.H.E
18.2k12070
$endgroup$
Hint:
$$fracfracdydtfracdxdt=fracdydx$$
answered 8 hours ago
E.H.EE.H.E
18.2k12070
answered 8 hours ago
E.H.EE.H.E
18.2k12070
answered 8 hours ago
E.H.EE.H.E
18.2k12070
answered 8 hours ago
E.H.EE.H.E
18.2k12070
18.2k12070
add a comment |
add a comment |
$begingroup$
We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.
answered 8 hours ago
auscryptauscrypt
4,326212
$endgroup$
add a comment |
$begingroup$
We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.
answered 8 hours ago
auscryptauscrypt
4,326212
$endgroup$
add a comment |
$begingroup$
We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.
answered 8 hours ago
auscryptauscrypt
4,326212
$endgroup$
We know that $doty=-x^3$. Divide this by $dotx=y$ and we get $fracdotydotx=frac-x^3y$. But $fracdotydotx$ is just $fracdydx$ so we're done.
answered 8 hours ago
auscryptauscrypt
4,326212
answered 8 hours ago
auscryptauscrypt
4,326212
answered 8 hours ago
auscryptauscrypt
4,326212
answered 8 hours ago
auscryptauscrypt
4,326212
4,326212
add a comment |
add a comment |
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2
$begingroup$
$fracdydt = fracdydxfracdxdt Rightarrow -x^3 = fracdydx y$
$endgroup$
– Cesareo
8 hours ago