How equation 6 has been solved to get equation 7? Can someone help me solving it with explanation?.












-1














I was reading one research paper, where they have solved the equation 6, and getting the answer as expressed in equation 7 . I can't understand how exactly they have solved it.
$$
frac{dC(t)}{dt} = alpha C u + frac{Q}{PBL} − alpha C(t)
tag{6}
$$

Assuming equilibrium, we can solve the differential eq 6. and obtain the following solution:
$$
C = Cu + frac{Q}{alpha PBL} + C_1 e^{−alpha t}tag{7}
$$

Thank you.










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  • It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
    – Axoren
    Dec 1 at 18:25
















-1














I was reading one research paper, where they have solved the equation 6, and getting the answer as expressed in equation 7 . I can't understand how exactly they have solved it.
$$
frac{dC(t)}{dt} = alpha C u + frac{Q}{PBL} − alpha C(t)
tag{6}
$$

Assuming equilibrium, we can solve the differential eq 6. and obtain the following solution:
$$
C = Cu + frac{Q}{alpha PBL} + C_1 e^{−alpha t}tag{7}
$$

Thank you.










share|cite|improve this question
























  • It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
    – Axoren
    Dec 1 at 18:25














-1












-1








-1







I was reading one research paper, where they have solved the equation 6, and getting the answer as expressed in equation 7 . I can't understand how exactly they have solved it.
$$
frac{dC(t)}{dt} = alpha C u + frac{Q}{PBL} − alpha C(t)
tag{6}
$$

Assuming equilibrium, we can solve the differential eq 6. and obtain the following solution:
$$
C = Cu + frac{Q}{alpha PBL} + C_1 e^{−alpha t}tag{7}
$$

Thank you.










share|cite|improve this question















I was reading one research paper, where they have solved the equation 6, and getting the answer as expressed in equation 7 . I can't understand how exactly they have solved it.
$$
frac{dC(t)}{dt} = alpha C u + frac{Q}{PBL} − alpha C(t)
tag{6}
$$

Assuming equilibrium, we can solve the differential eq 6. and obtain the following solution:
$$
C = Cu + frac{Q}{alpha PBL} + C_1 e^{−alpha t}tag{7}
$$

Thank you.







integration differential-equations






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edited Dec 1 at 18:42









rafa11111

1,121417




1,121417










asked Dec 1 at 18:18







user621794



















  • It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
    – Axoren
    Dec 1 at 18:25


















  • It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
    – Axoren
    Dec 1 at 18:25
















It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
– Axoren
Dec 1 at 18:25




It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
– Axoren
Dec 1 at 18:25










1 Answer
1






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oldest

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This is a separable equation, that can be written as
$$
frac{dC}{C - Cu-Q/alpha PBL} = -alpha dt
$$

Integrating both sides, you have
$$
ln |C-Cu - Q/alpha PBL| = C_0-alpha t,
$$

or
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    This is a separable equation, that can be written as
    $$
    frac{dC}{C - Cu-Q/alpha PBL} = -alpha dt
    $$

    Integrating both sides, you have
    $$
    ln |C-Cu - Q/alpha PBL| = C_0-alpha t,
    $$

    or
    $$
    C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
    $$






    share|cite|improve this answer


























      2














      This is a separable equation, that can be written as
      $$
      frac{dC}{C - Cu-Q/alpha PBL} = -alpha dt
      $$

      Integrating both sides, you have
      $$
      ln |C-Cu - Q/alpha PBL| = C_0-alpha t,
      $$

      or
      $$
      C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
      $$






      share|cite|improve this answer
























        2












        2








        2






        This is a separable equation, that can be written as
        $$
        frac{dC}{C - Cu-Q/alpha PBL} = -alpha dt
        $$

        Integrating both sides, you have
        $$
        ln |C-Cu - Q/alpha PBL| = C_0-alpha t,
        $$

        or
        $$
        C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
        $$






        share|cite|improve this answer












        This is a separable equation, that can be written as
        $$
        frac{dC}{C - Cu-Q/alpha PBL} = -alpha dt
        $$

        Integrating both sides, you have
        $$
        ln |C-Cu - Q/alpha PBL| = C_0-alpha t,
        $$

        or
        $$
        C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 18:28









        rafa11111

        1,121417




        1,121417






























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