Mixing
ODE with Euler Method
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I have to solve the following ODE: $y'(t)=y(t)+t$, $y(0)=0$
with Eulers Method in two steps, where $h=0.1$. I tried the following:
$y'(0)=y(0)+0=0$ and then I get $y(0.1)=y(0)+h*y'(0)=0$ but then everything will be 0. I do not get how to solve this.
ordinary-differential-equations eulers-method
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
$endgroup$
add a comment |
$begingroup$
I have to solve the following ODE: $y'(t)=y(t)+t$, $y(0)=0$
with Eulers Method in two steps, where $h=0.1$. I tried the following:
$y'(0)=y(0)+0=0$ and then I get $y(0.1)=y(0)+h*y'(0)=0$ but then everything will be 0. I do not get how to solve this.
ordinary-differential-equations eulers-method
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
$endgroup$
$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
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@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16
add a comment |
$begingroup$
I have to solve the following ODE: $y'(t)=y(t)+t$, $y(0)=0$
with Eulers Method in two steps, where $h=0.1$. I tried the following:
$y'(0)=y(0)+0=0$ and then I get $y(0.1)=y(0)+h*y'(0)=0$ but then everything will be 0. I do not get how to solve this.
ordinary-differential-equations eulers-method
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
$endgroup$
I have to solve the following ODE: $y'(t)=y(t)+t$, $y(0)=0$
with Eulers Method in two steps, where $h=0.1$. I tried the following:
$y'(0)=y(0)+0=0$ and then I get $y(0.1)=y(0)+h*y'(0)=0$ but then everything will be 0. I do not get how to solve this.
ordinary-differential-equations eulers-method
ordinary-differential-equations eulers-method
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
asked Jan 23 at 18:12
Natalie_94Natalie_94
316
316
$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
$begingroup$
@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16
add a comment |
$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
$begingroup$
@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16
$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
$begingroup$
@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16
add a comment |
1 Answer
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$begingroup$
In the next step you get $f(0.1, 0)=0.1$ so that you get a non-zero value for the approximation at $0.2$.
The leading error coefficient in $y_k=y(t_k)+c(t_k)h+O(h^2)$ is a solution of $$c'(t)=f_y(t,y)c-frac12y''=c-frac12e^t, \~~text{ so that }~~
(e^{-t}c)'=-frac12implies c(t)=-frac12te^t
$$
Plotting the actual error coefficient $frac{y_k-y(t_k)}h$ against this first term curve for several values of $h$ gives the image
So that indeed the Euler points are below the exact solution, for large $h$ this can be a large error, and for $h=0.1$ the error is as large as the value of the solution.
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
In the next step you get $f(0.1, 0)=0.1$ so that you get a non-zero value for the approximation at $0.2$.
The leading error coefficient in $y_k=y(t_k)+c(t_k)h+O(h^2)$ is a solution of $$c'(t)=f_y(t,y)c-frac12y''=c-frac12e^t, \~~text{ so that }~~
(e^{-t}c)'=-frac12implies c(t)=-frac12te^t
$$
Plotting the actual error coefficient $frac{y_k-y(t_k)}h$ against this first term curve for several values of $h$ gives the image
So that indeed the Euler points are below the exact solution, for large $h$ this can be a large error, and for $h=0.1$ the error is as large as the value of the solution.
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
In the next step you get $f(0.1, 0)=0.1$ so that you get a non-zero value for the approximation at $0.2$.
The leading error coefficient in $y_k=y(t_k)+c(t_k)h+O(h^2)$ is a solution of $$c'(t)=f_y(t,y)c-frac12y''=c-frac12e^t, \~~text{ so that }~~
(e^{-t}c)'=-frac12implies c(t)=-frac12te^t
$$
Plotting the actual error coefficient $frac{y_k-y(t_k)}h$ against this first term curve for several values of $h$ gives the image
So that indeed the Euler points are below the exact solution, for large $h$ this can be a large error, and for $h=0.1$ the error is as large as the value of the solution.
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
$endgroup$
add a comment |
$begingroup$
In the next step you get $f(0.1, 0)=0.1$ so that you get a non-zero value for the approximation at $0.2$.
The leading error coefficient in $y_k=y(t_k)+c(t_k)h+O(h^2)$ is a solution of $$c'(t)=f_y(t,y)c-frac12y''=c-frac12e^t, \~~text{ so that }~~
(e^{-t}c)'=-frac12implies c(t)=-frac12te^t
$$
Plotting the actual error coefficient $frac{y_k-y(t_k)}h$ against this first term curve for several values of $h$ gives the image
So that indeed the Euler points are below the exact solution, for large $h$ this can be a large error, and for $h=0.1$ the error is as large as the value of the solution.
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
$endgroup$
In the next step you get $f(0.1, 0)=0.1$ so that you get a non-zero value for the approximation at $0.2$.
The leading error coefficient in $y_k=y(t_k)+c(t_k)h+O(h^2)$ is a solution of $$c'(t)=f_y(t,y)c-frac12y''=c-frac12e^t, \~~text{ so that }~~
(e^{-t}c)'=-frac12implies c(t)=-frac12te^t
$$
Plotting the actual error coefficient $frac{y_k-y(t_k)}h$ against this first term curve for several values of $h$ gives the image
So that indeed the Euler points are below the exact solution, for large $h$ this can be a large error, and for $h=0.1$ the error is as large as the value of the solution.
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
edited Jan 29 at 18:01
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
answered Jan 23 at 19:15
LutzLLutzL
59.6k42057
59.6k42057
add a comment |
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$begingroup$
$y=0$ is the solution for initial condition $y(0)=0$. Choose a more interesting initial condition e.g. $y(0)=1$ and try again.
$endgroup$
– xidgel
Jan 23 at 18:56
$begingroup$
@xidgel : No, there is a non-zero forcing term, the general solution is $y(t)=ce^t-1-t$ and $c=1$ for $y(0)=0$.
$endgroup$
– LutzL
Jan 23 at 19:14
$begingroup$
@LutzL Thanks for catching that.
$endgroup$
– xidgel
Jan 23 at 19:16