Transformation of second order equation $y''-0.1y+2=0$ into system of first order equations [closed]
How do you convert this second order equation into a system of first order equations?
$$y''-0.1y+2=0$$
My own guess is the following:
If $u=y$ and $v=y'$, then $u'=y'$ and $v'=y''$, which gives
$$ u'=v text{and} v'=0.1u+2$$
calculus differential-equations
edited Dec 10 '18 at 16:34
rafa11111
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
closed as off-topic by Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers Dec 18 '18 at 20:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How do you convert this second order equation into a system of first order equations?
$$y''-0.1y+2=0$$
My own guess is the following:
If $u=y$ and $v=y'$, then $u'=y'$ and $v'=y''$, which gives
$$ u'=v text{and} v'=0.1u+2$$
calculus differential-equations
edited Dec 10 '18 at 16:34
rafa11111
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
closed as off-topic by Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers Dec 18 '18 at 20:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How do you convert this second order equation into a system of first order equations?
$$y''-0.1y+2=0$$
My own guess is the following:
If $u=y$ and $v=y'$, then $u'=y'$ and $v'=y''$, which gives
$$ u'=v text{and} v'=0.1u+2$$
calculus differential-equations
edited Dec 10 '18 at 16:34
rafa11111
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
How do you convert this second order equation into a system of first order equations?
$$y''-0.1y+2=0$$
My own guess is the following:
If $u=y$ and $v=y'$, then $u'=y'$ and $v'=y''$, which gives
$$ u'=v text{and} v'=0.1u+2$$
calculus differential-equations
calculus differential-equations
edited Dec 10 '18 at 16:34
rafa11111
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
edited Dec 10 '18 at 16:34
rafa11111
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
edited Dec 10 '18 at 16:34
rafa11111
1,116417
edited Dec 10 '18 at 16:34
rafa11111
1,116417
edited Dec 10 '18 at 16:34
rafa11111
1,116417
1,116417
asked Dec 10 '18 at 14:28
Kaare Lund
6
asked Dec 10 '18 at 14:28
Kaare Lund
6
asked Dec 10 '18 at 14:28
Kaare Lund
6
6
closed as off-topic by Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers Dec 18 '18 at 20:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers Dec 18 '18 at 20:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Brahadeesh, Nosrati, metamorphy, José Carlos Santos, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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oldest
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I will assume that $y$ is a function of $t$:
$$y''(t)=0.1y(t)-2$$
Let's call $y'(t)=v(t)$, which means that $y''(t)=v'(t)$:
$$v'(t)=0.1y(t)-2$$
And
$$y'(t)=v(t)$$
So your solution is almost correct, because you've just messed up a sign.
answered Dec 10 '18 at 14:52
Botond
5,5532732
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I will assume that $y$ is a function of $t$:
$$y''(t)=0.1y(t)-2$$
Let's call $y'(t)=v(t)$, which means that $y''(t)=v'(t)$:
$$v'(t)=0.1y(t)-2$$
And
$$y'(t)=v(t)$$
So your solution is almost correct, because you've just messed up a sign.
answered Dec 10 '18 at 14:52
Botond
5,5532732
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
add a comment |
I will assume that $y$ is a function of $t$:
$$y''(t)=0.1y(t)-2$$
Let's call $y'(t)=v(t)$, which means that $y''(t)=v'(t)$:
$$v'(t)=0.1y(t)-2$$
And
$$y'(t)=v(t)$$
So your solution is almost correct, because you've just messed up a sign.
answered Dec 10 '18 at 14:52
Botond
5,5532732
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
add a comment |
I will assume that $y$ is a function of $t$:
$$y''(t)=0.1y(t)-2$$
Let's call $y'(t)=v(t)$, which means that $y''(t)=v'(t)$:
$$v'(t)=0.1y(t)-2$$
And
$$y'(t)=v(t)$$
So your solution is almost correct, because you've just messed up a sign.
answered Dec 10 '18 at 14:52
Botond
5,5532732
I will assume that $y$ is a function of $t$:
$$y''(t)=0.1y(t)-2$$
Let's call $y'(t)=v(t)$, which means that $y''(t)=v'(t)$:
$$v'(t)=0.1y(t)-2$$
And
$$y'(t)=v(t)$$
So your solution is almost correct, because you've just messed up a sign.
answered Dec 10 '18 at 14:52
Botond
5,5532732
edited Dec 10 '18 at 16:02
answered Dec 10 '18 at 14:52
Botond
5,5532732
answered Dec 10 '18 at 14:52
Botond
5,5532732
answered Dec 10 '18 at 14:52
Botond
5,5532732
5,5532732
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
add a comment |
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
@LutzL yes, thank you!
– Botond
Dec 10 '18 at 15:35
add a comment |
