Integrating linearly
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I came across this question and just want to make sure my understanding is correct.
I need to find the general solution of:
$$
fracdxdt = a(1 - x)
$$
In this case, I'm finding the how $x$ changes with respect to $t$ so I'm integrating with respect to $t$. Does that mean the answer is $at - xat + C$?
Thanks :)
calculus integration
add a comment |Â
up vote
4
down vote
favorite
I came across this question and just want to make sure my understanding is correct.
I need to find the general solution of:
$$
fracdxdt = a(1 - x)
$$
In this case, I'm finding the how $x$ changes with respect to $t$ so I'm integrating with respect to $t$. Does that mean the answer is $at - xat + C$?
Thanks :)
calculus integration
2
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I came across this question and just want to make sure my understanding is correct.
I need to find the general solution of:
$$
fracdxdt = a(1 - x)
$$
In this case, I'm finding the how $x$ changes with respect to $t$ so I'm integrating with respect to $t$. Does that mean the answer is $at - xat + C$?
Thanks :)
calculus integration
I came across this question and just want to make sure my understanding is correct.
I need to find the general solution of:
$$
fracdxdt = a(1 - x)
$$
In this case, I'm finding the how $x$ changes with respect to $t$ so I'm integrating with respect to $t$. Does that mean the answer is $at - xat + C$?
Thanks :)
calculus integration
calculus integration
edited Aug 21 at 15:47
CulDeVu
15726
15726
asked Aug 21 at 9:23
Hews
515
515
2
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57
add a comment |Â
2
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57
2
2
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
8
down vote
You can't say that, simply because the function $x$ has a dependence in $t$ !
add a comment |Â
up vote
7
down vote
If you are given, for $x=x(t)$
$$fracdxdt = a(1-x)$$
Then, one can re-arrange (by abuse of notation), giving
beginalign
fracdx1-x &= a, dt \
implies int left(frac11-xright)dx &= a int dt
endalign
The right hand side is straightforward for you, the left requires some knowledge of standard integrals.
Does this help?
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
 |Â
show 1 more comment
up vote
3
down vote
You must write
$$fracdx1-x=a dt$$
add a comment |Â
up vote
2
down vote
If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $int tdt =frac t^22+C$. But if we use $int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $int xdt =t^2+C$, which is off by a factor of 2. Or if $x = sin(t)$, then we would have $int sin(t)dt=tsin(t)+C$ instead of $int sin(t)dt = cos(t)+C$. If we had that $int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.
Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^c_2t+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
You can't say that, simply because the function $x$ has a dependence in $t$ !
add a comment |Â
up vote
8
down vote
You can't say that, simply because the function $x$ has a dependence in $t$ !
add a comment |Â
up vote
8
down vote
up vote
8
down vote
You can't say that, simply because the function $x$ has a dependence in $t$ !
You can't say that, simply because the function $x$ has a dependence in $t$ !
answered Aug 21 at 9:26
tmaths
1,334113
1,334113
add a comment |Â
add a comment |Â
up vote
7
down vote
If you are given, for $x=x(t)$
$$fracdxdt = a(1-x)$$
Then, one can re-arrange (by abuse of notation), giving
beginalign
fracdx1-x &= a, dt \
implies int left(frac11-xright)dx &= a int dt
endalign
The right hand side is straightforward for you, the left requires some knowledge of standard integrals.
Does this help?
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
 |Â
show 1 more comment
up vote
7
down vote
If you are given, for $x=x(t)$
$$fracdxdt = a(1-x)$$
Then, one can re-arrange (by abuse of notation), giving
beginalign
fracdx1-x &= a, dt \
implies int left(frac11-xright)dx &= a int dt
endalign
The right hand side is straightforward for you, the left requires some knowledge of standard integrals.
Does this help?
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
 |Â
show 1 more comment
up vote
7
down vote
up vote
7
down vote
If you are given, for $x=x(t)$
$$fracdxdt = a(1-x)$$
Then, one can re-arrange (by abuse of notation), giving
beginalign
fracdx1-x &= a, dt \
implies int left(frac11-xright)dx &= a int dt
endalign
The right hand side is straightforward for you, the left requires some knowledge of standard integrals.
Does this help?
If you are given, for $x=x(t)$
$$fracdxdt = a(1-x)$$
Then, one can re-arrange (by abuse of notation), giving
beginalign
fracdx1-x &= a, dt \
implies int left(frac11-xright)dx &= a int dt
endalign
The right hand side is straightforward for you, the left requires some knowledge of standard integrals.
Does this help?
edited Aug 21 at 12:01
polfosol
5,36731843
5,36731843
answered Aug 21 at 9:28
Kevin
5,158722
5,158722
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
 |Â
show 1 more comment
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Awesome! Thanks for your help, does this lead to the answer "-ln(1-x) = at + C"?
â Hews
Aug 21 at 9:40
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
Yes. You probably want to turn it around and express $x$ in terms of $t$, and possibly rename the constant $e^-C$ to something else, as in this form has clearer meaning in the final function compared to just $C$.
â orion
Aug 21 at 9:45
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@orion not possibly, but necessarily. Even if we consider complex values of $C$, there is a value $e^-C$ canâÂÂt reach, namely 0, but the corresponding solution, the constant function $x=1$, is very much a valid solution of the original equation.
â Roman Odaisky
Aug 21 at 15:33
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
@Hews Yes that is exactly right, but as other posters point out, you possible would prefer to rearrange into $x= ldots$
â Kevin
Aug 22 at 13:41
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
So to express x in terms of t, does it end as x = e^(at)e(-C) - 1?. Thanks for your help everyone!
â Hews
Aug 22 at 23:25
 |Â
show 1 more comment
up vote
3
down vote
You must write
$$fracdx1-x=a dt$$
add a comment |Â
up vote
3
down vote
You must write
$$fracdx1-x=a dt$$
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You must write
$$fracdx1-x=a dt$$
You must write
$$fracdx1-x=a dt$$
answered Aug 21 at 9:28
Dr. Sonnhard Graubner
69k32761
69k32761
add a comment |Â
add a comment |Â
up vote
2
down vote
If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $int tdt =frac t^22+C$. But if we use $int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $int xdt =t^2+C$, which is off by a factor of 2. Or if $x = sin(t)$, then we would have $int sin(t)dt=tsin(t)+C$ instead of $int sin(t)dt = cos(t)+C$. If we had that $int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.
Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^c_2t+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.
add a comment |Â
up vote
2
down vote
If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $int tdt =frac t^22+C$. But if we use $int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $int xdt =t^2+C$, which is off by a factor of 2. Or if $x = sin(t)$, then we would have $int sin(t)dt=tsin(t)+C$ instead of $int sin(t)dt = cos(t)+C$. If we had that $int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.
Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^c_2t+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $int tdt =frac t^22+C$. But if we use $int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $int xdt =t^2+C$, which is off by a factor of 2. Or if $x = sin(t)$, then we would have $int sin(t)dt=tsin(t)+C$ instead of $int sin(t)dt = cos(t)+C$. If we had that $int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.
Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^c_2t+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.
If you write it as $dx = a(1-x)dt$ then it would be correct to integrate $x$ with respect to $t$. The trouble is, to perform the integration correctly, you would need to know what the dependence of $x$ on $t$ is, which is what you're trying to find out in the first place. To see what the problem with $int xdt =xt+C$ is, consider any function, e.g. $x=t$. If we substitute $t$ in for $x$ before integrating, we get $int tdt =frac t^22+C$. But if we use $int xdt =xt+C$ and substitute $t$ in for $x$ afterwards, we get $int xdt =t^2+C$, which is off by a factor of 2. Or if $x = sin(t)$, then we would have $int sin(t)dt=tsin(t)+C$ instead of $int sin(t)dt = cos(t)+C$. If we had that $int f(t)dt =tf(t)+C$, that would make the whole concept of an integral rather trivial; the integral of any function would just be that function times the independent variable. The identity $int xdt=xt+C$ works only if $x$ doesn't depend on $t$. Remember, an integral can be interpreted as the area under a curve. If $x$ is a constant, then we just have a rectangle with width $t$ and height $x$, so the area is $xt$. But if $x$ is varying with $t$, then we can't just take the value of $x$ at the end of the interval; clearly the area is going to depend on what $x$ is doing in between.
Note that if you get familiar with basic differential forms, you should get to a point where you recognize that when the derivative is proportional to the function value, you have an exponential function. In this case, the differential equation is modified by a constant term. So if you take the test solution $x = c_1e^c_2t+c_3$, solve for the derivative in terms of $c_1$, $c_2$, and $c_3$, and then plug that into the differential equation, then you can solve for $c_1$, $c_2$, and $c_3$. Note that one degree of freedom will remain, since this is a first-order equation and no initial condition is given.
edited Aug 21 at 15:25
answered Aug 21 at 15:20
Acccumulation
5,8442616
5,8442616
add a comment |Â
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2
You cannot integrate over $t$ if you have $x$ in the integral, which depends on $t$ in a yet unknown way. Separate the variables, it's the first elementary procedure you must learn to solve differential equations. Once you have $x$ on the left and $t$ on the right, you can proceed.
â orion
Aug 21 at 9:27
If $x(t) = t^2$, for example, then $int x, dt$ is clearly not $tx$.
â anomaly
Aug 21 at 19:57