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Solve by a series of powers around the given ordinary point
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Solve by a series of powers around the given ordinary point, $$ x^2y'' + xy'-y = 0 hspace5mmx_0 = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=sum_n=0^inftyc_n(x-2)^n\y'=sum_n=1^inftyc_nn(x-2)^n-1\y''=sum_n=2^inftyc_nn(n-1)(x-2)^n-2$$
calculus differential-equations power-series
asked Jul 27 at 21:36
Santiago Seeker
577
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up vote
1
down vote
favorite
Solve by a series of powers around the given ordinary point, $$ x^2y'' + xy'-y = 0 hspace5mmx_0 = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=sum_n=0^inftyc_n(x-2)^n\y'=sum_n=1^inftyc_nn(x-2)^n-1\y''=sum_n=2^inftyc_nn(n-1)(x-2)^n-2$$
calculus differential-equations power-series
asked Jul 27 at 21:36
Santiago Seeker
577
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Solve by a series of powers around the given ordinary point, $$ x^2y'' + xy'-y = 0 hspace5mmx_0 = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=sum_n=0^inftyc_n(x-2)^n\y'=sum_n=1^inftyc_nn(x-2)^n-1\y''=sum_n=2^inftyc_nn(n-1)(x-2)^n-2$$
calculus differential-equations power-series
asked Jul 27 at 21:36
Santiago Seeker
577
Solve by a series of powers around the given ordinary point, $$ x^2y'' + xy'-y = 0 hspace5mmx_0 = 2 $$
I have problems to continue with the development, I start with the assumption
$$y=sum_n=0^inftyc_n(x-2)^n\y'=sum_n=1^inftyc_nn(x-2)^n-1\y''=sum_n=2^inftyc_nn(n-1)(x-2)^n-2$$
calculus differential-equations power-series
asked Jul 27 at 21:36
Santiago Seeker
577
edited Jul 27 at 22:00
asked Jul 27 at 21:36
Santiago Seeker
577
asked Jul 27 at 21:36
Santiago Seeker
577
asked Jul 27 at 21:36
Santiago Seeker
577
577
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Hint.
Note that $frac 1x$ is a solution for the DE so try something with
$$
y = frac 12sum_k=0^infty(-1)^kfrac(x-2)^k2^k+sum_k=0^inftya_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\
x^2 = (x-2)^2+4(x-2)+4
$$
answered Jul 27 at 22:51
Cesareo
5,6412412
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
add a comment |Â
up vote
0
down vote
Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
beginalign
y &= sum_n=0^infty c_nt^n \ \
(t+2)y' &= (t+2)sum_n=1^infty c_nnt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=1^infty 2nc_nt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=0^infty 2(n+1)c_n+1t^n \ \
(t+2)^2y'' &= (t^2+4t+4)sum_n=2^infty n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=2^infty 4n(n-1)c_nt^n-1 + sum_n=2^infty 4n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=1^infty 4(n+1)nc_n+1t^n + sum_n=0^infty 4(n+2)(n+1)c_n+2t^n
endalign
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + sum_n=2^infty big[4(n+2)(n+1)c_n+2 + 2(n+1)(2n+1)c_n+1 + (n^2-1)c_nt^n big] $$
Then we have the recurring relations
beginalign
c_2 &= frac18(c_0-2c_1) \
c_3 &= -frac12c_2 \
c_n+2 &= frac2n+12(n+2)c_n+1 + fracn-14(n+2)c_n
endalign
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + fracbx $$
answered Jul 28 at 7:18
Dylan
11.4k31026
add a comment |Â
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0
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Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2+frac1X,D-frac1X^2=left(D+frac2Xright),left(D-frac1Xright),.$$
First, suppose $z:=left(D-dfrac1Xright)y$. Then, $left(D+dfrac2Xright),z=0$, or equivalently, $D,left(X^2,zright)=0$. Consequently,
$$z(x)=-frac2ax^2text for some constant a,.$$
Now, $left(D-dfrac1Xright),y=z$ implies that $D,left(dfrac1X,yright)=dfrac1X,z$. Therefore, for some constant $b$, we have
$$y(x)=x,int,frac1x,left(-frac2ax^2right),textdx=x,Biggl(aleft(frac1x^2right)+bBiggr)=a,left(frac1xright)+b,x,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b,x$ for some constant $b$.
I should have realized that there is an easier decomposition
$$D^2+frac1X,D-frac1X^2=D,left(D+frac1Xright),.$$
This leads to a very simple solution. Note that
$left(D^2+dfrac1X,D-dfrac1X^2right),y=0$ implies
$$D,Biggl(frac1X,D,(X,y)Biggr)=0,.$$
Thus, $big(D,(X,y)big)(x)=2A,x$ for some constant $A$, and
$$y(x)=frac1x,left(A,x^2+Bright)=A,x+B,left(frac1xright)$$
for some constant $B$.
answered Aug 2 at 8:38
Batominovski
23k22777
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint.
Note that $frac 1x$ is a solution for the DE so try something with
$$
y = frac 12sum_k=0^infty(-1)^kfrac(x-2)^k2^k+sum_k=0^inftya_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\
x^2 = (x-2)^2+4(x-2)+4
$$
answered Jul 27 at 22:51
Cesareo
5,6412412
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
add a comment |Â
up vote
1
down vote
accepted
Hint.
Note that $frac 1x$ is a solution for the DE so try something with
$$
y = frac 12sum_k=0^infty(-1)^kfrac(x-2)^k2^k+sum_k=0^inftya_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\
x^2 = (x-2)^2+4(x-2)+4
$$
answered Jul 27 at 22:51
Cesareo
5,6412412
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint.
Note that $frac 1x$ is a solution for the DE so try something with
$$
y = frac 12sum_k=0^infty(-1)^kfrac(x-2)^k2^k+sum_k=0^inftya_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\
x^2 = (x-2)^2+4(x-2)+4
$$
answered Jul 27 at 22:51
Cesareo
5,6412412
Hint.
Note that $frac 1x$ is a solution for the DE so try something with
$$
y = frac 12sum_k=0^infty(-1)^kfrac(x-2)^k2^k+sum_k=0^inftya_k (x-2)^k
$$
note also that
$$
x = (x-2) + 2\
x^2 = (x-2)^2+4(x-2)+4
$$
answered Jul 27 at 22:51
Cesareo
5,6412412
edited Jul 28 at 0:05
answered Jul 27 at 22:51
Cesareo
5,6412412
answered Jul 27 at 22:51
Cesareo
5,6412412
answered Jul 27 at 22:51
Cesareo
5,6412412
5,6412412
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
add a comment |Â
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
but I have problems when replacing, for example when it is around an ordinary point $ x_0 = 0 $ then the assumption is useful, $$ y = sum_n = 0^infty c_nx^n $$ but everything remains in function of $ x ^ n $ to obtain the equation of recurrence. I'm stuck there
– Santiago Seeker
Jul 27 at 22:57
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
@SantiagoSeeker I included a note that may helps.
– Cesareo
Jul 28 at 0:07
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
Ohh!!...right!! ...thank you...
– Santiago Seeker
Jul 28 at 0:16
add a comment |Â
up vote
0
down vote
Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
beginalign
y &= sum_n=0^infty c_nt^n \ \
(t+2)y' &= (t+2)sum_n=1^infty c_nnt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=1^infty 2nc_nt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=0^infty 2(n+1)c_n+1t^n \ \
(t+2)^2y'' &= (t^2+4t+4)sum_n=2^infty n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=2^infty 4n(n-1)c_nt^n-1 + sum_n=2^infty 4n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=1^infty 4(n+1)nc_n+1t^n + sum_n=0^infty 4(n+2)(n+1)c_n+2t^n
endalign
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + sum_n=2^infty big[4(n+2)(n+1)c_n+2 + 2(n+1)(2n+1)c_n+1 + (n^2-1)c_nt^n big] $$
Then we have the recurring relations
beginalign
c_2 &= frac18(c_0-2c_1) \
c_3 &= -frac12c_2 \
c_n+2 &= frac2n+12(n+2)c_n+1 + fracn-14(n+2)c_n
endalign
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + fracbx $$
answered Jul 28 at 7:18
Dylan
11.4k31026
add a comment |Â
up vote
0
down vote
Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
beginalign
y &= sum_n=0^infty c_nt^n \ \
(t+2)y' &= (t+2)sum_n=1^infty c_nnt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=1^infty 2nc_nt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=0^infty 2(n+1)c_n+1t^n \ \
(t+2)^2y'' &= (t^2+4t+4)sum_n=2^infty n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=2^infty 4n(n-1)c_nt^n-1 + sum_n=2^infty 4n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=1^infty 4(n+1)nc_n+1t^n + sum_n=0^infty 4(n+2)(n+1)c_n+2t^n
endalign
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + sum_n=2^infty big[4(n+2)(n+1)c_n+2 + 2(n+1)(2n+1)c_n+1 + (n^2-1)c_nt^n big] $$
Then we have the recurring relations
beginalign
c_2 &= frac18(c_0-2c_1) \
c_3 &= -frac12c_2 \
c_n+2 &= frac2n+12(n+2)c_n+1 + fracn-14(n+2)c_n
endalign
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + fracbx $$
answered Jul 28 at 7:18
Dylan
11.4k31026
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
beginalign
y &= sum_n=0^infty c_nt^n \ \
(t+2)y' &= (t+2)sum_n=1^infty c_nnt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=1^infty 2nc_nt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=0^infty 2(n+1)c_n+1t^n \ \
(t+2)^2y'' &= (t^2+4t+4)sum_n=2^infty n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=2^infty 4n(n-1)c_nt^n-1 + sum_n=2^infty 4n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=1^infty 4(n+1)nc_n+1t^n + sum_n=0^infty 4(n+2)(n+1)c_n+2t^n
endalign
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + sum_n=2^infty big[4(n+2)(n+1)c_n+2 + 2(n+1)(2n+1)c_n+1 + (n^2-1)c_nt^n big] $$
Then we have the recurring relations
beginalign
c_2 &= frac18(c_0-2c_1) \
c_3 &= -frac12c_2 \
c_n+2 &= frac2n+12(n+2)c_n+1 + fracn-14(n+2)c_n
endalign
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + fracbx $$
answered Jul 28 at 7:18
Dylan
11.4k31026
Let $t=x-2$, then the equation becomes
$$ (t+2)^2y'' + (t+2)y' - y = 0 $$
Simplifying each term at a time:
beginalign
y &= sum_n=0^infty c_nt^n \ \
(t+2)y' &= (t+2)sum_n=1^infty c_nnt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=1^infty 2nc_nt^n-1 \
&= sum_n=1^infty nc_nt^n + sum_n=0^infty 2(n+1)c_n+1t^n \ \
(t+2)^2y'' &= (t^2+4t+4)sum_n=2^infty n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=2^infty 4n(n-1)c_nt^n-1 + sum_n=2^infty 4n(n-1)c_nt^n-2 \
&= sum_n=2^infty n(n-1)c_nt^n + sum_n=1^infty 4(n+1)nc_n+1t^n + sum_n=0^infty 4(n+2)(n+1)c_n+2t^n
endalign
Combining terms:
$$ (8c_2+2c_1-c_0) + (24c_3 + 12c_2)t + sum_n=2^infty big[4(n+2)(n+1)c_n+2 + 2(n+1)(2n+1)c_n+1 + (n^2-1)c_nt^n big] $$
Then we have the recurring relations
beginalign
c_2 &= frac18(c_0-2c_1) \
c_3 &= -frac12c_2 \
c_n+2 &= frac2n+12(n+2)c_n+1 + fracn-14(n+2)c_n
endalign
Finding a closed form (in terms of $c_0$ and $c_1$) is a bit more involved.
Using the Cauchy-Euler method, you can obtain the "official" solution:
$$ y(x) = ax + fracbx $$
answered Jul 28 at 7:18
Dylan
11.4k31026
edited Jul 28 at 7:25
answered Jul 28 at 7:18
Dylan
11.4k31026
answered Jul 28 at 7:18
Dylan
11.4k31026
answered Jul 28 at 7:18
Dylan
11.4k31026
11.4k31026
add a comment |Â
add a comment |Â
up vote
0
down vote
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2+frac1X,D-frac1X^2=left(D+frac2Xright),left(D-frac1Xright),.$$
First, suppose $z:=left(D-dfrac1Xright)y$. Then, $left(D+dfrac2Xright),z=0$, or equivalently, $D,left(X^2,zright)=0$. Consequently,
$$z(x)=-frac2ax^2text for some constant a,.$$
Now, $left(D-dfrac1Xright),y=z$ implies that $D,left(dfrac1X,yright)=dfrac1X,z$. Therefore, for some constant $b$, we have
$$y(x)=x,int,frac1x,left(-frac2ax^2right),textdx=x,Biggl(aleft(frac1x^2right)+bBiggr)=a,left(frac1xright)+b,x,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b,x$ for some constant $b$.
I should have realized that there is an easier decomposition
$$D^2+frac1X,D-frac1X^2=D,left(D+frac1Xright),.$$
This leads to a very simple solution. Note that
$left(D^2+dfrac1X,D-dfrac1X^2right),y=0$ implies
$$D,Biggl(frac1X,D,(X,y)Biggr)=0,.$$
Thus, $big(D,(X,y)big)(x)=2A,x$ for some constant $A$, and
$$y(x)=frac1x,left(A,x^2+Bright)=A,x+B,left(frac1xright)$$
for some constant $B$.
answered Aug 2 at 8:38
Batominovski
23k22777
add a comment |Â
up vote
0
down vote
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2+frac1X,D-frac1X^2=left(D+frac2Xright),left(D-frac1Xright),.$$
First, suppose $z:=left(D-dfrac1Xright)y$. Then, $left(D+dfrac2Xright),z=0$, or equivalently, $D,left(X^2,zright)=0$. Consequently,
$$z(x)=-frac2ax^2text for some constant a,.$$
Now, $left(D-dfrac1Xright),y=z$ implies that $D,left(dfrac1X,yright)=dfrac1X,z$. Therefore, for some constant $b$, we have
$$y(x)=x,int,frac1x,left(-frac2ax^2right),textdx=x,Biggl(aleft(frac1x^2right)+bBiggr)=a,left(frac1xright)+b,x,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b,x$ for some constant $b$.
I should have realized that there is an easier decomposition
$$D^2+frac1X,D-frac1X^2=D,left(D+frac1Xright),.$$
This leads to a very simple solution. Note that
$left(D^2+dfrac1X,D-dfrac1X^2right),y=0$ implies
$$D,Biggl(frac1X,D,(X,y)Biggr)=0,.$$
Thus, $big(D,(X,y)big)(x)=2A,x$ for some constant $A$, and
$$y(x)=frac1x,left(A,x^2+Bright)=A,x+B,left(frac1xright)$$
for some constant $B$.
answered Aug 2 at 8:38
Batominovski
23k22777
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2+frac1X,D-frac1X^2=left(D+frac2Xright),left(D-frac1Xright),.$$
First, suppose $z:=left(D-dfrac1Xright)y$. Then, $left(D+dfrac2Xright),z=0$, or equivalently, $D,left(X^2,zright)=0$. Consequently,
$$z(x)=-frac2ax^2text for some constant a,.$$
Now, $left(D-dfrac1Xright),y=z$ implies that $D,left(dfrac1X,yright)=dfrac1X,z$. Therefore, for some constant $b$, we have
$$y(x)=x,int,frac1x,left(-frac2ax^2right),textdx=x,Biggl(aleft(frac1x^2right)+bBiggr)=a,left(frac1xright)+b,x,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b,x$ for some constant $b$.
I should have realized that there is an easier decomposition
$$D^2+frac1X,D-frac1X^2=D,left(D+frac1Xright),.$$
This leads to a very simple solution. Note that
$left(D^2+dfrac1X,D-dfrac1X^2right),y=0$ implies
$$D,Biggl(frac1X,D,(X,y)Biggr)=0,.$$
Thus, $big(D,(X,y)big)(x)=2A,x$ for some constant $A$, and
$$y(x)=frac1x,left(A,x^2+Bright)=A,x+B,left(frac1xright)$$
for some constant $B$.
answered Aug 2 at 8:38
Batominovski
23k22777
Define the operators $D$ and $X$ by $(D,h)(x):=h'(x)$ and $(X,h)(x):=x,h(x)$. For any function $phi$, we also define the operator $phi(X)$ to be $big(phi(X),hbig)(x):=phi(x),h(x)$. Note that
$$D^2+frac1X,D-frac1X^2=left(D+frac2Xright),left(D-frac1Xright),.$$
First, suppose $z:=left(D-dfrac1Xright)y$. Then, $left(D+dfrac2Xright),z=0$, or equivalently, $D,left(X^2,zright)=0$. Consequently,
$$z(x)=-frac2ax^2text for some constant a,.$$
Now, $left(D-dfrac1Xright),y=z$ implies that $D,left(dfrac1X,yright)=dfrac1X,z$. Therefore, for some constant $b$, we have
$$y(x)=x,int,frac1x,left(-frac2ax^2right),textdx=x,Biggl(aleft(frac1x^2right)+bBiggr)=a,left(frac1xright)+b,x,.$$
Thus, the only power-series solution to this differential equation is $y(x)=b,x$ for some constant $b$.
I should have realized that there is an easier decomposition
$$D^2+frac1X,D-frac1X^2=D,left(D+frac1Xright),.$$
This leads to a very simple solution. Note that
$left(D^2+dfrac1X,D-dfrac1X^2right),y=0$ implies
$$D,Biggl(frac1X,D,(X,y)Biggr)=0,.$$
Thus, $big(D,(X,y)big)(x)=2A,x$ for some constant $A$, and
$$y(x)=frac1x,left(A,x^2+Bright)=A,x+B,left(frac1xright)$$
for some constant $B$.
answered Aug 2 at 8:38
Batominovski
23k22777
edited Aug 2 at 9:17
answered Aug 2 at 8:38
Batominovski
23k22777
answered Aug 2 at 8:38
Batominovski
23k22777
answered Aug 2 at 8:38
Batominovski
23k22777
23k22777
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