Mixing
Exercise: first-order linear differential equation
$begingroup$
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
asked Jan 19 at 19:31
user3204810user3204810
1947
$endgroup$
add a comment |
$begingroup$
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
asked Jan 19 at 19:31
user3204810user3204810
1947
$endgroup$
$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
1
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37
add a comment |
$begingroup$
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
asked Jan 19 at 19:31
user3204810user3204810
1947
$endgroup$
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
asked Jan 19 at 19:31
user3204810user3204810
1947
asked Jan 19 at 19:31
user3204810user3204810
1947
edited Jan 19 at 19:44
user3204810
asked Jan 19 at 19:31
user3204810user3204810
1947
asked Jan 19 at 19:31
user3204810user3204810
1947
asked Jan 19 at 19:31
user3204810user3204810
1947
1947
$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
1
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37
add a comment |
$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
1
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37
$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
1
1
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $$C$$
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
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
$$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$C$$
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
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 $$C$$
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
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 $$C$$
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
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 $$C$$
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 19:39
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
76.7k42866
add a comment |
add a comment |
$begingroup$
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
$$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
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
$$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
$endgroup$
add a comment |
$begingroup$
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
$$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
$endgroup$
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
$$y(x)=y_0e^{-kx}+frac pkRightarrow y(0)=y_0+frac pkcolor{red}neq y_0$$
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
answered Jan 19 at 19:48
mrtaurhomrtaurho
5,59051440
5,59051440
add a comment |
add a comment |
$begingroup$
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.
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
add a comment |
$begingroup$
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.
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
$endgroup$
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
add a comment |
$begingroup$
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.
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
$endgroup$
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.
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
answered Jan 19 at 19:44
Rhys HughesRhys Hughes
6,9441530
6,9441530
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
add a comment |
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
$begingroup$
I corrected. Sorry for the mistake
$endgroup$
– user3204810
Jan 19 at 19:56
add a comment |
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$begingroup$
Could you maybe add the initial condition? Without one cannot really determine which solution is the right one.
$endgroup$
– mrtaurho
Jan 19 at 19:38
1
$begingroup$
When $x=0$ your equation reads $y(0)=y(0)+p/k$
$endgroup$
– user121049
Jan 19 at 19:39
$begingroup$
my teacher's file says "for $x in [0,X]$", so I think the initial condition is $y(0) = 0$
$endgroup$
– user3204810
Jan 19 at 19:43
$begingroup$
Both are right but in your solution the constant is not $y(0)$
$endgroup$
– Dylan
Jan 20 at 6:37