Exercise: first-order linear differential equation












1












$begingroup$


This is a first-order linear differential equation:



$$y' = -ky + p$$



where $k$ and $p$ are constant. Based on my calculations, the solution is



$$y(x) = y(0) cdot exp^{-kx} + ; dfrac{p}{k}$$



while my teacher's file says



$$y(x) = dfrac{p}{k} + left[y(0) - dfrac{p}{k} right] exp^{-kx}$$



Which one is the right one?



Thank you in advance






ordinary-differential-equations







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$endgroup$












  • $begingroup$
    Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
    $endgroup$
    – mrtaurho
    1 hour ago










  • $begingroup$
    When $x=0$ your equation reads $y(0)=y(0)+p/k$
    $endgroup$
    – user121049
    1 hour ago










  • $begingroup$
    my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
    $endgroup$
    – user3204810
    1 hour ago
















1












$begingroup$


This is a first-order linear differential equation:



$$y' = -ky + p$$



where $k$ and $p$ are constant. Based on my calculations, the solution is



$$y(x) = y(0) cdot exp^{-kx} + ; dfrac{p}{k}$$



while my teacher's file says



$$y(x) = dfrac{p}{k} + left[y(0) - dfrac{p}{k} right] exp^{-kx}$$



Which one is the right one?



Thank you in advance






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
    $endgroup$
    – mrtaurho
    1 hour ago










  • $begingroup$
    When $x=0$ your equation reads $y(0)=y(0)+p/k$
    $endgroup$
    – user121049
    1 hour ago










  • $begingroup$
    my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
    $endgroup$
    – user3204810
    1 hour ago














1












1








1





$begingroup$


This is a first-order linear differential equation:



$$y' = -ky + p$$



where $k$ and $p$ are constant. Based on my calculations, the solution is



$$y(x) = y(0) cdot exp^{-kx} + ; dfrac{p}{k}$$



while my teacher's file says



$$y(x) = dfrac{p}{k} + left[y(0) - dfrac{p}{k} right] exp^{-kx}$$



Which one is the right one?



Thank you in advance






ordinary-differential-equations







share|cite|improve this question











$endgroup$




This is a first-order linear differential equation:



$$y' = -ky + p$$



where $k$ and $p$ are constant. Based on my calculations, the solution is



$$y(x) = y(0) cdot exp^{-kx} + ; dfrac{p}{k}$$



while my teacher's file says



$$y(x) = dfrac{p}{k} + left[y(0) - dfrac{p}{k} right] exp^{-kx}$$



Which one is the right one?



Thank you in advance






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







user3204810

















asked 1 hour ago









user3204810user3204810

1947




1947












  • $begingroup$
    Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
    $endgroup$
    – mrtaurho
    1 hour ago










  • $begingroup$
    When $x=0$ your equation reads $y(0)=y(0)+p/k$
    $endgroup$
    – user121049
    1 hour ago










  • $begingroup$
    my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
    $endgroup$
    – user3204810
    1 hour ago


















  • $begingroup$
    Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
    $endgroup$
    – mrtaurho
    1 hour ago










  • $begingroup$
    When $x=0$ your equation reads $y(0)=y(0)+p/k$
    $endgroup$
    – user121049
    1 hour ago










  • $begingroup$
    my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
    $endgroup$
    – user3204810
    1 hour ago
















$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
1 hour ago




$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
1 hour ago












$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
1 hour ago




$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
1 hour ago












$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
1 hour ago




$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
1 hour ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

The general solution is given by $$y(x)=frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=frac{p}{k}+C$$ to compute $$C$$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following



    begin{align}
    y'&=-ky+p\
    y'&=-kleft(y-frac pkright)\
    frac{y'}{y-frac pk}&=-k\
    int frac{mathrm dy}{y-frac pk}&=-kintmathrm dx\
    logleft(y-frac pkright)&=-kx+c\
    y-frac pk&=ce^{-kx}\
    end{align}




    $$therefore~y(x)~=~ce^{-kx}+frac pk$$




    Determining the constant $c$ by using $y(0)=y_0$ we get



    $$y(0)=c+frac pk=y_0Rightarrow~c=y_0-frac pk$$




    $$therefore~y(x)=frac pk+left[y_0-frac pkright]e^{-kx}$$




    I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since



    $$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Firstly, this is first-order.



      We let $$y'+ky=p$$
      Then use an integrating factor:



      $$IF=e^{int k dx}=e^{kx}$$



      Then the trick:
      $$ycdot IF =int{IFcdot RHS dx}$$
      $$to ye^{kx}=int{pe^{kx} dx}$$
      $$to y=frac pk +Ce^{-kx}$$



      You appear correct therefore, unless the initial condition changes something.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I corrected. Sorry for the mistake
        $endgroup$
        – user3204810
        1 hour ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The general solution is given by $$y(x)=frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=frac{p}{k}+C$$ to compute $$C$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The general solution is given by $$y(x)=frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=frac{p}{k}+C$$ to compute $$C$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The general solution is given by $$y(x)=frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=frac{p}{k}+C$$ to compute $$C$$






          share|cite|improve this answer









          $endgroup$



          The general solution is given by $$y(x)=frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=frac{p}{k}+C$$ to compute $$C$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          74k42865




          74k42865























              2












              $begingroup$

              The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following



              begin{align}
              y'&=-ky+p\
              y'&=-kleft(y-frac pkright)\
              frac{y'}{y-frac pk}&=-k\
              int frac{mathrm dy}{y-frac pk}&=-kintmathrm dx\
              logleft(y-frac pkright)&=-kx+c\
              y-frac pk&=ce^{-kx}\
              end{align}




              $$therefore~y(x)~=~ce^{-kx}+frac pk$$




              Determining the constant $c$ by using $y(0)=y_0$ we get



              $$y(0)=c+frac pk=y_0Rightarrow~c=y_0-frac pk$$




              $$therefore~y(x)=frac pk+left[y_0-frac pkright]e^{-kx}$$




              I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since



              $$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following



                begin{align}
                y'&=-ky+p\
                y'&=-kleft(y-frac pkright)\
                frac{y'}{y-frac pk}&=-k\
                int frac{mathrm dy}{y-frac pk}&=-kintmathrm dx\
                logleft(y-frac pkright)&=-kx+c\
                y-frac pk&=ce^{-kx}\
                end{align}




                $$therefore~y(x)~=~ce^{-kx}+frac pk$$




                Determining the constant $c$ by using $y(0)=y_0$ we get



                $$y(0)=c+frac pk=y_0Rightarrow~c=y_0-frac pk$$




                $$therefore~y(x)=frac pk+left[y_0-frac pkright]e^{-kx}$$




                I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since



                $$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following



                  begin{align}
                  y'&=-ky+p\
                  y'&=-kleft(y-frac pkright)\
                  frac{y'}{y-frac pk}&=-k\
                  int frac{mathrm dy}{y-frac pk}&=-kintmathrm dx\
                  logleft(y-frac pkright)&=-kx+c\
                  y-frac pk&=ce^{-kx}\
                  end{align}




                  $$therefore~y(x)~=~ce^{-kx}+frac pk$$




                  Determining the constant $c$ by using $y(0)=y_0$ we get



                  $$y(0)=c+frac pk=y_0Rightarrow~c=y_0-frac pk$$




                  $$therefore~y(x)=frac pk+left[y_0-frac pkright]e^{-kx}$$




                  I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since



                  $$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$






                  share|cite|improve this answer









                  $endgroup$



                  The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following



                  begin{align}
                  y'&=-ky+p\
                  y'&=-kleft(y-frac pkright)\
                  frac{y'}{y-frac pk}&=-k\
                  int frac{mathrm dy}{y-frac pk}&=-kintmathrm dx\
                  logleft(y-frac pkright)&=-kx+c\
                  y-frac pk&=ce^{-kx}\
                  end{align}




                  $$therefore~y(x)~=~ce^{-kx}+frac pk$$




                  Determining the constant $c$ by using $y(0)=y_0$ we get



                  $$y(0)=c+frac pk=y_0Rightarrow~c=y_0-frac pk$$




                  $$therefore~y(x)=frac pk+left[y_0-frac pkright]e^{-kx}$$




                  I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since



                  $$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  mrtaurhomrtaurho

                  4,15121234




                  4,15121234























                      1












                      $begingroup$

                      Firstly, this is first-order.



                      We let $$y'+ky=p$$
                      Then use an integrating factor:



                      $$IF=e^{int k dx}=e^{kx}$$



                      Then the trick:
                      $$ycdot IF =int{IFcdot RHS dx}$$
                      $$to ye^{kx}=int{pe^{kx} dx}$$
                      $$to y=frac pk +Ce^{-kx}$$



                      You appear correct therefore, unless the initial condition changes something.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I corrected. Sorry for the mistake
                        $endgroup$
                        – user3204810
                        1 hour ago
















                      1












                      $begingroup$

                      Firstly, this is first-order.



                      We let $$y'+ky=p$$
                      Then use an integrating factor:



                      $$IF=e^{int k dx}=e^{kx}$$



                      Then the trick:
                      $$ycdot IF =int{IFcdot RHS dx}$$
                      $$to ye^{kx}=int{pe^{kx} dx}$$
                      $$to y=frac pk +Ce^{-kx}$$



                      You appear correct therefore, unless the initial condition changes something.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I corrected. Sorry for the mistake
                        $endgroup$
                        – user3204810
                        1 hour ago














                      1












                      1








                      1





                      $begingroup$

                      Firstly, this is first-order.



                      We let $$y'+ky=p$$
                      Then use an integrating factor:



                      $$IF=e^{int k dx}=e^{kx}$$



                      Then the trick:
                      $$ycdot IF =int{IFcdot RHS dx}$$
                      $$to ye^{kx}=int{pe^{kx} dx}$$
                      $$to y=frac pk +Ce^{-kx}$$



                      You appear correct therefore, unless the initial condition changes something.






                      share|cite|improve this answer









                      $endgroup$



                      Firstly, this is first-order.



                      We let $$y'+ky=p$$
                      Then use an integrating factor:



                      $$IF=e^{int k dx}=e^{kx}$$



                      Then the trick:
                      $$ycdot IF =int{IFcdot RHS dx}$$
                      $$to ye^{kx}=int{pe^{kx} dx}$$
                      $$to y=frac pk +Ce^{-kx}$$



                      You appear correct therefore, unless the initial condition changes something.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Rhys HughesRhys Hughes

                      5,5561528




                      5,5561528












                      • $begingroup$
                        I corrected. Sorry for the mistake
                        $endgroup$
                        – user3204810
                        1 hour ago


















                      • $begingroup$
                        I corrected. Sorry for the mistake
                        $endgroup$
                        – user3204810
                        1 hour ago
















                      $begingroup$
                      I corrected. Sorry for the mistake
                      $endgroup$
                      – user3204810
                      1 hour ago




                      $begingroup$
                      I corrected. Sorry for the mistake
                      $endgroup$
                      – user3204810
                      1 hour ago


















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