How equation 6 has been solved to get equation 7? Can someone help me solving it with explanation?.
$begingroup$
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 ordinary-differential-equations
edited Dec 1 '18 at 18:42
rafa11111
1,119417
$endgroup$
add a comment |
$begingroup$
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 ordinary-differential-equations
edited Dec 1 '18 at 18:42
rafa11111
1,119417
$endgroup$
$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25
add a comment |
$begingroup$
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 ordinary-differential-equations
edited Dec 1 '18 at 18:42
rafa11111
1,119417
$endgroup$
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 ordinary-differential-equations
integration ordinary-differential-equations
edited Dec 1 '18 at 18:42
rafa11111
1,119417
edited Dec 1 '18 at 18:42
rafa11111
1,119417
edited Dec 1 '18 at 18:42
rafa11111
1,119417
edited Dec 1 '18 at 18:42
rafa11111
1,119417
edited Dec 1 '18 at 18:42
rafa11111
1,119417
1,119417
asked Dec 1 '18 at 18:18
user621794
$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25
add a comment |
$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25
$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25
$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
$endgroup$
add a comment |
$begingroup$
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
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
$endgroup$
add a comment |
$begingroup$
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
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
$endgroup$
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
$$
C = Cu + frac{Q}{alpha PBL} C_1 e^{-alpha t}.
$$
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
answered Dec 1 '18 at 18:28
rafa11111rafa11111
1,119417
1,119417
add a comment |
add a comment |
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$begingroup$
It may be helpful to include the paper as there may be an explanation in the text that could help answerers help you.
$endgroup$
– Axoren
Dec 1 '18 at 18:25