differential equation - beginner question [closed]












1














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 $c_2 = frac{1}{c_1}$?










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closed as off-topic by RRL, Saad, Cesareo, metamorphy, José Carlos Santos Dec 21 '18 at 12:20


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If this question can be reworded to fit the rules in the help center, please edit the question.


















    1














    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 $c_2 = frac{1}{c_1}$?










    share|cite|improve this 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.
















      1












      1








      1







      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 $c_2 = frac{1}{c_1}$?










      share|cite|improve this question















      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 differential-equations






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 16 '18 at 15:45







      gariban17

















      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.






















          1 Answer
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          6














          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 $1/c$.






          share|cite|improve this answer




























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6














            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 $1/c$.






            share|cite|improve this answer


























              6














              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 $1/c$.






              share|cite|improve this answer
























                6












                6








                6






                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 $1/c$.






                share|cite|improve this answer












                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 $1/c$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 9:40









                RebellosRebellos

                14.5k31246




                14.5k31246















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