Mixing
Solving $(D^5-D)y = 8sin x$ using operator methods
$begingroup$
This question is similar to this link but here it involves $sin x$ and it creates a problem. I tried writing $sin x = Im (e^{ix})$ but that doesnt help:
$$(D^5-D) y = 8Im (e^{ix})$$
Now $(D^5-D)$ has factor $D=i$ so we need to factor that out and apply it on right side:
$$(D-1)(D+i)(D+1)D y=frac{1}{D-i}(Im(e^{ix}))\
implies y = frac{8}{(i-1)(i+1)(2i)(i)}Im(e^{ix}int e^{ix} e^{-ix} dx)$$
which gives $y_p = 2xsin x$. Another method would have been assuming $y_p = x(Acos x + Bsin x)$ but that too would have been lengthy
Is the method in this answer correct, because it is using complex numbers loosely. Is this operator method?
calculus ordinary-differential-equations
asked Jan 20 at 12:16
jeeajeea
60715
$endgroup$
add a comment |
$begingroup$
This question is similar to this link but here it involves $sin x$ and it creates a problem. I tried writing $sin x = Im (e^{ix})$ but that doesnt help:
$$(D^5-D) y = 8Im (e^{ix})$$
Now $(D^5-D)$ has factor $D=i$ so we need to factor that out and apply it on right side:
$$(D-1)(D+i)(D+1)D y=frac{1}{D-i}(Im(e^{ix}))\
implies y = frac{8}{(i-1)(i+1)(2i)(i)}Im(e^{ix}int e^{ix} e^{-ix} dx)$$
which gives $y_p = 2xsin x$. Another method would have been assuming $y_p = x(Acos x + Bsin x)$ but that too would have been lengthy
Is the method in this answer correct, because it is using complex numbers loosely. Is this operator method?
calculus ordinary-differential-equations
asked Jan 20 at 12:16
jeeajeea
60715
$endgroup$
add a comment |
$begingroup$
This question is similar to this link but here it involves $sin x$ and it creates a problem. I tried writing $sin x = Im (e^{ix})$ but that doesnt help:
$$(D^5-D) y = 8Im (e^{ix})$$
Now $(D^5-D)$ has factor $D=i$ so we need to factor that out and apply it on right side:
$$(D-1)(D+i)(D+1)D y=frac{1}{D-i}(Im(e^{ix}))\
implies y = frac{8}{(i-1)(i+1)(2i)(i)}Im(e^{ix}int e^{ix} e^{-ix} dx)$$
which gives $y_p = 2xsin x$. Another method would have been assuming $y_p = x(Acos x + Bsin x)$ but that too would have been lengthy
Is the method in this answer correct, because it is using complex numbers loosely. Is this operator method?
calculus ordinary-differential-equations
asked Jan 20 at 12:16
jeeajeea
60715
$endgroup$
This question is similar to this link but here it involves $sin x$ and it creates a problem. I tried writing $sin x = Im (e^{ix})$ but that doesnt help:
$$(D^5-D) y = 8Im (e^{ix})$$
Now $(D^5-D)$ has factor $D=i$ so we need to factor that out and apply it on right side:
$$(D-1)(D+i)(D+1)D y=frac{1}{D-i}(Im(e^{ix}))\
implies y = frac{8}{(i-1)(i+1)(2i)(i)}Im(e^{ix}int e^{ix} e^{-ix} dx)$$
which gives $y_p = 2xsin x$. Another method would have been assuming $y_p = x(Acos x + Bsin x)$ but that too would have been lengthy
Is the method in this answer correct, because it is using complex numbers loosely. Is this operator method?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 20 at 12:16
jeeajeea
60715
asked Jan 20 at 12:16
jeeajeea
60715
asked Jan 20 at 12:16
jeeajeea
60715
asked Jan 20 at 12:16
jeeajeea
60715
asked Jan 20 at 12:16
jeeajeea
60715
60715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can use the same method suggested in 1. Let us find a solution $y$ of the equation
$$
(D^5-D)[y]=8e^{ix}
$$ first. Write $T=D^5-D$. For any $rinBbb C$, it holds that
$$
T[e^{rx}]=D^5[e^{rx}]-D[e^{rx}]=(r^5-r)e^{rx}.
$$ Plugging $r=i$ into the equation does not give the solution because $i^5-i=0$. Now, by differentiating both sides with respect to $r$, we get
$$
frac{partial}{partial r}T[e^{rx}]=T[frac{partial}{partial r}e^{rx}]=T[xe^{rx}]=frac{partial}{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
$$
Plugging $r=i$ into the expression, we get
$$
T[xe^{ix}]=(5i^4-1)e^{ix}+(i^5-i)xe^{ix}=4e^{ix}.
$$ This shows $T[2xe^{ix}]=8e^{ix}$. By separating real and imaginary parts, we get
$$
T[2xcos x]=8cos x,quad T[2xsin x]=8sin x.
$$ This allows us a particular solution $y(x) =2xsin x$ satisfying $T[y]=8sin x$.
answered Jan 20 at 12:32
SongSong
16.5k1741
$endgroup$
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
You can use the same method suggested in 1. Let us find a solution $y$ of the equation
$$
(D^5-D)[y]=8e^{ix}
$$ first. Write $T=D^5-D$. For any $rinBbb C$, it holds that
$$
T[e^{rx}]=D^5[e^{rx}]-D[e^{rx}]=(r^5-r)e^{rx}.
$$ Plugging $r=i$ into the equation does not give the solution because $i^5-i=0$. Now, by differentiating both sides with respect to $r$, we get
$$
frac{partial}{partial r}T[e^{rx}]=T[frac{partial}{partial r}e^{rx}]=T[xe^{rx}]=frac{partial}{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
$$
Plugging $r=i$ into the expression, we get
$$
T[xe^{ix}]=(5i^4-1)e^{ix}+(i^5-i)xe^{ix}=4e^{ix}.
$$ This shows $T[2xe^{ix}]=8e^{ix}$. By separating real and imaginary parts, we get
$$
T[2xcos x]=8cos x,quad T[2xsin x]=8sin x.
$$ This allows us a particular solution $y(x) =2xsin x$ satisfying $T[y]=8sin x$.
answered Jan 20 at 12:32
SongSong
16.5k1741
$endgroup$
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
add a comment |
$begingroup$
You can use the same method suggested in 1. Let us find a solution $y$ of the equation
$$
(D^5-D)[y]=8e^{ix}
$$ first. Write $T=D^5-D$. For any $rinBbb C$, it holds that
$$
T[e^{rx}]=D^5[e^{rx}]-D[e^{rx}]=(r^5-r)e^{rx}.
$$ Plugging $r=i$ into the equation does not give the solution because $i^5-i=0$. Now, by differentiating both sides with respect to $r$, we get
$$
frac{partial}{partial r}T[e^{rx}]=T[frac{partial}{partial r}e^{rx}]=T[xe^{rx}]=frac{partial}{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
$$
Plugging $r=i$ into the expression, we get
$$
T[xe^{ix}]=(5i^4-1)e^{ix}+(i^5-i)xe^{ix}=4e^{ix}.
$$ This shows $T[2xe^{ix}]=8e^{ix}$. By separating real and imaginary parts, we get
$$
T[2xcos x]=8cos x,quad T[2xsin x]=8sin x.
$$ This allows us a particular solution $y(x) =2xsin x$ satisfying $T[y]=8sin x$.
answered Jan 20 at 12:32
SongSong
16.5k1741
$endgroup$
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
add a comment |
$begingroup$
You can use the same method suggested in 1. Let us find a solution $y$ of the equation
$$
(D^5-D)[y]=8e^{ix}
$$ first. Write $T=D^5-D$. For any $rinBbb C$, it holds that
$$
T[e^{rx}]=D^5[e^{rx}]-D[e^{rx}]=(r^5-r)e^{rx}.
$$ Plugging $r=i$ into the equation does not give the solution because $i^5-i=0$. Now, by differentiating both sides with respect to $r$, we get
$$
frac{partial}{partial r}T[e^{rx}]=T[frac{partial}{partial r}e^{rx}]=T[xe^{rx}]=frac{partial}{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
$$
Plugging $r=i$ into the expression, we get
$$
T[xe^{ix}]=(5i^4-1)e^{ix}+(i^5-i)xe^{ix}=4e^{ix}.
$$ This shows $T[2xe^{ix}]=8e^{ix}$. By separating real and imaginary parts, we get
$$
T[2xcos x]=8cos x,quad T[2xsin x]=8sin x.
$$ This allows us a particular solution $y(x) =2xsin x$ satisfying $T[y]=8sin x$.
answered Jan 20 at 12:32
SongSong
16.5k1741
$endgroup$
You can use the same method suggested in 1. Let us find a solution $y$ of the equation
$$
(D^5-D)[y]=8e^{ix}
$$ first. Write $T=D^5-D$. For any $rinBbb C$, it holds that
$$
T[e^{rx}]=D^5[e^{rx}]-D[e^{rx}]=(r^5-r)e^{rx}.
$$ Plugging $r=i$ into the equation does not give the solution because $i^5-i=0$. Now, by differentiating both sides with respect to $r$, we get
$$
frac{partial}{partial r}T[e^{rx}]=T[frac{partial}{partial r}e^{rx}]=T[xe^{rx}]=frac{partial}{partial r}left((r^5-r)e^{rx}right)=(5r^4-1)e^{rx}+(r^5-r)xe^{rx}.
$$
Plugging $r=i$ into the expression, we get
$$
T[xe^{ix}]=(5i^4-1)e^{ix}+(i^5-i)xe^{ix}=4e^{ix}.
$$ This shows $T[2xe^{ix}]=8e^{ix}$. By separating real and imaginary parts, we get
$$
T[2xcos x]=8cos x,quad T[2xsin x]=8sin x.
$$ This allows us a particular solution $y(x) =2xsin x$ satisfying $T[y]=8sin x$.
answered Jan 20 at 12:32
SongSong
16.5k1741
answered Jan 20 at 12:32
SongSong
16.5k1741
answered Jan 20 at 12:32
SongSong
16.5k1741
answered Jan 20 at 12:32
SongSong
16.5k1741
16.5k1741
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
add a comment |
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
THanks a lot again! is the method I put in the question also fine
$endgroup$
– jeea
Jan 20 at 12:40
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
$begingroup$
@jeea It came close to being right ... Note that to invert $(D-i)$, i.e. to get $(D-i)^{-1}[f]$, we should calculate the convolution$$int_0^x e^{i(x-t)}f(t)dt .$$ In this case, $$int_0^x e^{i(x-t)}e^{it}dt =int_0^x e^{ix}dt = xe^{ix}.$$
$endgroup$
– Song
Jan 20 at 12:45
add a comment |
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