differential equation - beginner question [closed]
$begingroup$
If I have a differential equation on the form
$$y = y' cdot c_1$$
can I freely solve for $y'$ and use the solution for
$$y' = y cdot c_2$$
where $c_2 = frac{1}{c_1}$?
calculus ordinary-differential-equations
asked Dec 16 '18 at 9:37
gariban17gariban17
284
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closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
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." – RRL, Saad, Cesareo, José Carlos Santos
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$begingroup$
If I have a differential equation on the form
$$y = y' cdot c_1$$
can I freely solve for $y'$ and use the solution for
$$y' = y cdot c_2$$
where $c_2 = frac{1}{c_1}$?
calculus ordinary-differential-equations
asked Dec 16 '18 at 9:37
gariban17gariban17
284
$endgroup$
closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
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." – RRL, Saad, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If I have a differential equation on the form
$$y = y' cdot c_1$$
can I freely solve for $y'$ and use the solution for
$$y' = y cdot c_2$$
where $c_2 = frac{1}{c_1}$?
calculus ordinary-differential-equations
asked Dec 16 '18 at 9:37
gariban17gariban17
284
$endgroup$
If I have a differential equation on the form
$$y = y' cdot c_1$$
can I freely solve for $y'$ and use the solution for
$$y' = y cdot c_2$$
where $c_2 = frac{1}{c_1}$?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Dec 16 '18 at 9:37
gariban17gariban17
284
asked Dec 16 '18 at 9:37
gariban17gariban17
284
edited Dec 16 '18 at 15:45
gariban17
asked Dec 16 '18 at 9:37
gariban17gariban17
284
asked Dec 16 '18 at 9:37
gariban17gariban17
284
asked Dec 16 '18 at 9:37
gariban17gariban17
284
284
closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
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." – RRL, Saad, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos Santos Dec 21 '18 at 12:20
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." – RRL, Saad, Cesareo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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Yes, of course. Assuming that $c in mathbb R$ is a constant, then if $c neq 0$ :
$$y = y' cdot c Leftrightarrow y' = y cdot frac{1}{c} equiv y cdot c$$
Since $c$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $c$. Note that only if you have some certain restrictions for $c$, then you will need to take these in mind on how they affect the expression $1/c$.
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, of course. Assuming that $c in mathbb R$ is a constant, then if $c neq 0$ :
$$y = y' cdot c Leftrightarrow y' = y cdot frac{1}{c} equiv y cdot c$$
Since $c$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $c$. Note that only if you have some certain restrictions for $c$, then you will need to take these in mind on how they affect the expression $1/c$.
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
Yes, of course. Assuming that $c in mathbb R$ is a constant, then if $c neq 0$ :
$$y = y' cdot c Leftrightarrow y' = y cdot frac{1}{c} equiv y cdot c$$
Since $c$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $c$. Note that only if you have some certain restrictions for $c$, then you will need to take these in mind on how they affect the expression $1/c$.
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
$endgroup$
add a comment |
$begingroup$
Yes, of course. Assuming that $c in mathbb R$ is a constant, then if $c neq 0$ :
$$y = y' cdot c Leftrightarrow y' = y cdot frac{1}{c} equiv y cdot c$$
Since $c$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $c$. Note that only if you have some certain restrictions for $c$, then you will need to take these in mind on how they affect the expression $1/c$.
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
$endgroup$
Yes, of course. Assuming that $c in mathbb R$ is a constant, then if $c neq 0$ :
$$y = y' cdot c Leftrightarrow y' = y cdot frac{1}{c} equiv y cdot c$$
Since $c$ is an arbitrary constant, any expression of it will also be a constant, so you can always "manipulate" it to be just $c$. Note that only if you have some certain restrictions for $c$, then you will need to take these in mind on how they affect the expression $1/c$.
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
answered Dec 16 '18 at 9:40
RebellosRebellos
14.5k31246
14.5k31246
add a comment |
add a comment |
