Solve by means of regular perturbation to obtain an approximate solution up to and including...
$begingroup$
I have to solve the following ODEs:
$y''+ y = e^{epsilon cos x}$, $y(0,epsilon)=y(1,epsilon)=0$;
$y''+ y = epsilon y^2$, $y(0,epsilon)=1$, $y'(0,epsilon)=0$.
I am having trouble with these two questions.
For the first one, I got:
begin{align}
y''_0 + y_0 &=1,& y_0(0)&=y_0(1)=0\
y''_1 + y_1 &= cos x,& y_1(0)&=y_1(1)=0
end{align}
Not sure where the $cos x$ is from.
For the second question, unsure how $y_0(x)=cos x$
and $y_1(x)=1/2-1/3(cos x)-1/6(cos 2x)$
numerical-methods perturbation-theory
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
$endgroup$
add a comment |
$begingroup$
I have to solve the following ODEs:
$y''+ y = e^{epsilon cos x}$, $y(0,epsilon)=y(1,epsilon)=0$;
$y''+ y = epsilon y^2$, $y(0,epsilon)=1$, $y'(0,epsilon)=0$.
I am having trouble with these two questions.
For the first one, I got:
begin{align}
y''_0 + y_0 &=1,& y_0(0)&=y_0(1)=0\
y''_1 + y_1 &= cos x,& y_1(0)&=y_1(1)=0
end{align}
Not sure where the $cos x$ is from.
For the second question, unsure how $y_0(x)=cos x$
and $y_1(x)=1/2-1/3(cos x)-1/6(cos 2x)$
numerical-methods perturbation-theory
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
$endgroup$
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01
add a comment |
$begingroup$
I have to solve the following ODEs:
$y''+ y = e^{epsilon cos x}$, $y(0,epsilon)=y(1,epsilon)=0$;
$y''+ y = epsilon y^2$, $y(0,epsilon)=1$, $y'(0,epsilon)=0$.
I am having trouble with these two questions.
For the first one, I got:
begin{align}
y''_0 + y_0 &=1,& y_0(0)&=y_0(1)=0\
y''_1 + y_1 &= cos x,& y_1(0)&=y_1(1)=0
end{align}
Not sure where the $cos x$ is from.
For the second question, unsure how $y_0(x)=cos x$
and $y_1(x)=1/2-1/3(cos x)-1/6(cos 2x)$
numerical-methods perturbation-theory
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
$endgroup$
I have to solve the following ODEs:
$y''+ y = e^{epsilon cos x}$, $y(0,epsilon)=y(1,epsilon)=0$;
$y''+ y = epsilon y^2$, $y(0,epsilon)=1$, $y'(0,epsilon)=0$.
I am having trouble with these two questions.
For the first one, I got:
begin{align}
y''_0 + y_0 &=1,& y_0(0)&=y_0(1)=0\
y''_1 + y_1 &= cos x,& y_1(0)&=y_1(1)=0
end{align}
Not sure where the $cos x$ is from.
For the second question, unsure how $y_0(x)=cos x$
and $y_1(x)=1/2-1/3(cos x)-1/6(cos 2x)$
numerical-methods perturbation-theory
numerical-methods perturbation-theory
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
edited Dec 19 '18 at 18:50
LutzL
59.1k42056
59.1k42056
asked Dec 18 '18 at 23:25
luffyluffy
92
asked Dec 18 '18 at 23:25
luffyluffy
92
asked Dec 18 '18 at 23:25
luffyluffy
92
92
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01
add a comment |
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01
$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function
$$
e^{ϵcos x}=1+ϵcos x+frac{ϵ^2}2cos^2 x+frac{ϵ^3}6cos^2 x+...
$$) The initial equation is the harmonic oscillator
$$
y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0
$$
which has $y_0(x)=cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is
$$
y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0
$$
and $cos^2x$ is as is well-known equal to $frac12(1+cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function
$$
e^{ϵcos x}=1+ϵcos x+frac{ϵ^2}2cos^2 x+frac{ϵ^3}6cos^2 x+...
$$) The initial equation is the harmonic oscillator
$$
y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0
$$
which has $y_0(x)=cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is
$$
y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0
$$
and $cos^2x$ is as is well-known equal to $frac12(1+cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
add a comment |
$begingroup$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function
$$
e^{ϵcos x}=1+ϵcos x+frac{ϵ^2}2cos^2 x+frac{ϵ^3}6cos^2 x+...
$$) The initial equation is the harmonic oscillator
$$
y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0
$$
which has $y_0(x)=cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is
$$
y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0
$$
and $cos^2x$ is as is well-known equal to $frac12(1+cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
$endgroup$
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
add a comment |
$begingroup$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function
$$
e^{ϵcos x}=1+ϵcos x+frac{ϵ^2}2cos^2 x+frac{ϵ^3}6cos^2 x+...
$$) The initial equation is the harmonic oscillator
$$
y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0
$$
which has $y_0(x)=cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is
$$
y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0
$$
and $cos^2x$ is as is well-known equal to $frac12(1+cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
$endgroup$
) You can write the right side as series expansion in $ϵ$, that is, insert the series for the exponential function
$$
e^{ϵcos x}=1+ϵcos x+frac{ϵ^2}2cos^2 x+frac{ϵ^3}6cos^2 x+...
$$) The initial equation is the harmonic oscillator
$$
y_0''+y_0=0,~~y_0(0)=1,~~y_0'(0)=0
$$
which has $y_0(x)=cos(x)$ as solution. The next equation of the terms of first order in $ϵ$ is
$$
y_1''+y_1=y_0^2,~~y_1(0)=1,~~y_1'(0)=0
$$
and $cos^2x$ is as is well-known equal to $frac12(1+cos(2x))$. Apply the method of undetermined coefficients to find the corresponding particular solution.
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
answered Dec 19 '18 at 9:11
LutzLLutzL
59.1k42056
59.1k42056
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
add a comment |
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
solving y''+y=0 gives me: c1cos(x)+c2sin(x) and solving for the particular integral: substituting acos(x)+bsin(x) back into the ODE: y= acos(x)+bsin(x) y'=-asin(x)+bcos(x) y''=-acos(x)-bsin(x) plugging this into the ODE would just cancel and leave 0=cos(x) So, how does the solution give y=1/2sin(x)
$endgroup$
– luffy
Dec 19 '18 at 17:35
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
$begingroup$
Now apply variation of constants or undetermined coefficients to find a particular solution for the right side $1$.
$endgroup$
– LutzL
Dec 19 '18 at 18:48
add a comment |
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$begingroup$
Welcome to MSE. Please read this text about how to ask a good question.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 0:01