Simple system of Linear ODEs
I have this simple system of ODEs for rates $r_1,r_2$ and $a(t), b(t)$:
$$a' = r_2b - r_1a$$
$$b' = -r_2b + r_1a$$
I am trying to solve for $a(t), b(t)$, but I am not sure what I am doing wrong. I first start with differentiating the first ODE:
$a'' = r_2b'-r_1a'$
$; ; ; ;$ $ = r_2[-r_2b + r_1a] - r_1a'$
$; ; ; ;$ $ = r_2[(-a'-r_1a) + r_1a] - r_1a'$
$; ; ; ;$ $ = -r_2a' - r_1a'$
$; ; ; ;$ $ = a'(-r_2-r_1)$
$$rightarrow a(t) = a_0e^{(-r_2-r_1)t}$$
Following the same logic I also get:
$$rightarrow b(t) = b_0e^{(-r_2-r_1)t}$$
However I am confused as to why both $a$ and $b$ have the same decay rates. I might be wrong, but I would have assumed that the solutions would have been
$$a(t) = a_0e^{(r_2-r_1)t}$$
$$b(t) = b_0e^{(r_1-r_2)t}$$
Where did I go wrong?
calculus differential-equations dynamical-systems
edited Dec 11 '18 at 14:19
LutzL
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
add a comment |
I have this simple system of ODEs for rates $r_1,r_2$ and $a(t), b(t)$:
$$a' = r_2b - r_1a$$
$$b' = -r_2b + r_1a$$
I am trying to solve for $a(t), b(t)$, but I am not sure what I am doing wrong. I first start with differentiating the first ODE:
$a'' = r_2b'-r_1a'$
$; ; ; ;$ $ = r_2[-r_2b + r_1a] - r_1a'$
$; ; ; ;$ $ = r_2[(-a'-r_1a) + r_1a] - r_1a'$
$; ; ; ;$ $ = -r_2a' - r_1a'$
$; ; ; ;$ $ = a'(-r_2-r_1)$
$$rightarrow a(t) = a_0e^{(-r_2-r_1)t}$$
Following the same logic I also get:
$$rightarrow b(t) = b_0e^{(-r_2-r_1)t}$$
However I am confused as to why both $a$ and $b$ have the same decay rates. I might be wrong, but I would have assumed that the solutions would have been
$$a(t) = a_0e^{(r_2-r_1)t}$$
$$b(t) = b_0e^{(r_1-r_2)t}$$
Where did I go wrong?
calculus differential-equations dynamical-systems
edited Dec 11 '18 at 14:19
LutzL
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
2
Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45
add a comment |
I have this simple system of ODEs for rates $r_1,r_2$ and $a(t), b(t)$:
$$a' = r_2b - r_1a$$
$$b' = -r_2b + r_1a$$
I am trying to solve for $a(t), b(t)$, but I am not sure what I am doing wrong. I first start with differentiating the first ODE:
$a'' = r_2b'-r_1a'$
$; ; ; ;$ $ = r_2[-r_2b + r_1a] - r_1a'$
$; ; ; ;$ $ = r_2[(-a'-r_1a) + r_1a] - r_1a'$
$; ; ; ;$ $ = -r_2a' - r_1a'$
$; ; ; ;$ $ = a'(-r_2-r_1)$
$$rightarrow a(t) = a_0e^{(-r_2-r_1)t}$$
Following the same logic I also get:
$$rightarrow b(t) = b_0e^{(-r_2-r_1)t}$$
However I am confused as to why both $a$ and $b$ have the same decay rates. I might be wrong, but I would have assumed that the solutions would have been
$$a(t) = a_0e^{(r_2-r_1)t}$$
$$b(t) = b_0e^{(r_1-r_2)t}$$
Where did I go wrong?
calculus differential-equations dynamical-systems
edited Dec 11 '18 at 14:19
LutzL
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
I have this simple system of ODEs for rates $r_1,r_2$ and $a(t), b(t)$:
$$a' = r_2b - r_1a$$
$$b' = -r_2b + r_1a$$
I am trying to solve for $a(t), b(t)$, but I am not sure what I am doing wrong. I first start with differentiating the first ODE:
$a'' = r_2b'-r_1a'$
$; ; ; ;$ $ = r_2[-r_2b + r_1a] - r_1a'$
$; ; ; ;$ $ = r_2[(-a'-r_1a) + r_1a] - r_1a'$
$; ; ; ;$ $ = -r_2a' - r_1a'$
$; ; ; ;$ $ = a'(-r_2-r_1)$
$$rightarrow a(t) = a_0e^{(-r_2-r_1)t}$$
Following the same logic I also get:
$$rightarrow b(t) = b_0e^{(-r_2-r_1)t}$$
However I am confused as to why both $a$ and $b$ have the same decay rates. I might be wrong, but I would have assumed that the solutions would have been
$$a(t) = a_0e^{(r_2-r_1)t}$$
$$b(t) = b_0e^{(r_1-r_2)t}$$
Where did I go wrong?
calculus differential-equations dynamical-systems
calculus differential-equations dynamical-systems
edited Dec 11 '18 at 14:19
LutzL
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
edited Dec 11 '18 at 14:19
LutzL
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
edited Dec 11 '18 at 14:19
LutzL
56k42054
edited Dec 11 '18 at 14:19
LutzL
56k42054
edited Dec 11 '18 at 14:19
LutzL
56k42054
56k42054
asked Dec 9 '18 at 20:07
JBL
463210
asked Dec 9 '18 at 20:07
JBL
463210
asked Dec 9 '18 at 20:07
JBL
463210
463210
2
Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45
add a comment |
2
Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45
2
2
Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45
Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45
add a comment |
2 Answers
2
active
oldest
votes
Hint:
Notice that $a' = -b'$
With this in mind, look again in your calculations step by step.
You will find a mistake and you will answer your own question
answered Dec 9 '18 at 20:32
Bo5man
517
add a comment |
The conclusion from $$a''=-ra',$$ $r=r_1+r_2$, is not $a(t)=a_0e^{-rt}.$ As a second order linear differential equation it has two basis solutions and two integration constants.
Instead, you should get
$$
a(t)=a_0+c(e^{-rt}-1).
$$
Insert into $b=(a'+r_1a)/r_2$ to get
$$
b(t)=frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)implies b_0=frac{r_1a_0-rc}{r_2}
$$
so that
$$
c=frac{r_1a_0-r_2b_0}{r}.
$$
answered Dec 9 '18 at 20:50
LutzL
56k42054
add a comment |
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
votes
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active
oldest
votes
Hint:
Notice that $a' = -b'$
With this in mind, look again in your calculations step by step.
You will find a mistake and you will answer your own question
answered Dec 9 '18 at 20:32
Bo5man
517
add a comment |
Hint:
Notice that $a' = -b'$
With this in mind, look again in your calculations step by step.
You will find a mistake and you will answer your own question
answered Dec 9 '18 at 20:32
Bo5man
517
add a comment |
Hint:
Notice that $a' = -b'$
With this in mind, look again in your calculations step by step.
You will find a mistake and you will answer your own question
answered Dec 9 '18 at 20:32
Bo5man
517
Hint:
Notice that $a' = -b'$
With this in mind, look again in your calculations step by step.
You will find a mistake and you will answer your own question
answered Dec 9 '18 at 20:32
Bo5man
517
answered Dec 9 '18 at 20:32
Bo5man
517
answered Dec 9 '18 at 20:32
Bo5man
517
answered Dec 9 '18 at 20:32
Bo5man
517
517
add a comment |
add a comment |
The conclusion from $$a''=-ra',$$ $r=r_1+r_2$, is not $a(t)=a_0e^{-rt}.$ As a second order linear differential equation it has two basis solutions and two integration constants.
Instead, you should get
$$
a(t)=a_0+c(e^{-rt}-1).
$$
Insert into $b=(a'+r_1a)/r_2$ to get
$$
b(t)=frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)implies b_0=frac{r_1a_0-rc}{r_2}
$$
so that
$$
c=frac{r_1a_0-r_2b_0}{r}.
$$
answered Dec 9 '18 at 20:50
LutzL
56k42054
add a comment |
The conclusion from $$a''=-ra',$$ $r=r_1+r_2$, is not $a(t)=a_0e^{-rt}.$ As a second order linear differential equation it has two basis solutions and two integration constants.
Instead, you should get
$$
a(t)=a_0+c(e^{-rt}-1).
$$
Insert into $b=(a'+r_1a)/r_2$ to get
$$
b(t)=frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)implies b_0=frac{r_1a_0-rc}{r_2}
$$
so that
$$
c=frac{r_1a_0-r_2b_0}{r}.
$$
answered Dec 9 '18 at 20:50
LutzL
56k42054
add a comment |
The conclusion from $$a''=-ra',$$ $r=r_1+r_2$, is not $a(t)=a_0e^{-rt}.$ As a second order linear differential equation it has two basis solutions and two integration constants.
Instead, you should get
$$
a(t)=a_0+c(e^{-rt}-1).
$$
Insert into $b=(a'+r_1a)/r_2$ to get
$$
b(t)=frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)implies b_0=frac{r_1a_0-rc}{r_2}
$$
so that
$$
c=frac{r_1a_0-r_2b_0}{r}.
$$
answered Dec 9 '18 at 20:50
LutzL
56k42054
The conclusion from $$a''=-ra',$$ $r=r_1+r_2$, is not $a(t)=a_0e^{-rt}.$ As a second order linear differential equation it has two basis solutions and two integration constants.
Instead, you should get
$$
a(t)=a_0+c(e^{-rt}-1).
$$
Insert into $b=(a'+r_1a)/r_2$ to get
$$
b(t)=frac{r_1(a_0-c)-r_2ce^{-rt}}{r_2}=b_0-c(e^{-rt}-1)implies b_0=frac{r_1a_0-rc}{r_2}
$$
so that
$$
c=frac{r_1a_0-r_2b_0}{r}.
$$
answered Dec 9 '18 at 20:50
LutzL
56k42054
answered Dec 9 '18 at 20:50
LutzL
56k42054
answered Dec 9 '18 at 20:50
LutzL
56k42054
answered Dec 9 '18 at 20:50
LutzL
56k42054
56k42054
add a comment |
add a comment |
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Don't you think your last edit removed important information from the question?
– rafa11111
Dec 10 '18 at 19:45