Mixing
Reduction of Order via Substitution
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Suppose $u_1=sinx^2$ is a solution of
$$xu''-u'+4x^3u=0Rightarrow u''-x^-1u'+4x^2u=0 (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_1intfracdxu^2_1textexpleft(int p(x) dxright) , p(x)=-x^-1$$
I have found that $$u_2=fracsinx^22int textcosec^2x dx=-fraccosx^22$$
Now I tried to replicate this using the substitution $u_2=u_1v(x)$.
After finding the first and second derivatives of this equation and substituting into $(1)$, I get
$$v''sinx^2+v'left(4xcosx^2-x^-1sinx^2right)=0$$
Letting $w=v'$,
$$fracdwdxsinx^2+wleft(4xcosx^2-x^-1sinx^2right)=0$$
I tried to simplify this using the integrating factor
$$textexpleft(int 4xcosx^2-x^-1sinx^2 dxright)$$ but could not compute it.
How can I solve this problem using my suggested substitution?
calculus differential-equations reduction-of-order-ode
asked Jul 29 at 6:44
Bell
565113
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up vote
2
down vote
favorite
Suppose $u_1=sinx^2$ is a solution of
$$xu''-u'+4x^3u=0Rightarrow u''-x^-1u'+4x^2u=0 (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_1intfracdxu^2_1textexpleft(int p(x) dxright) , p(x)=-x^-1$$
I have found that $$u_2=fracsinx^22int textcosec^2x dx=-fraccosx^22$$
Now I tried to replicate this using the substitution $u_2=u_1v(x)$.
After finding the first and second derivatives of this equation and substituting into $(1)$, I get
$$v''sinx^2+v'left(4xcosx^2-x^-1sinx^2right)=0$$
Letting $w=v'$,
$$fracdwdxsinx^2+wleft(4xcosx^2-x^-1sinx^2right)=0$$
I tried to simplify this using the integrating factor
$$textexpleft(int 4xcosx^2-x^-1sinx^2 dxright)$$ but could not compute it.
How can I solve this problem using my suggested substitution?
calculus differential-equations reduction-of-order-ode
asked Jul 29 at 6:44
Bell
565113
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $u_1=sinx^2$ is a solution of
$$xu''-u'+4x^3u=0Rightarrow u''-x^-1u'+4x^2u=0 (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_1intfracdxu^2_1textexpleft(int p(x) dxright) , p(x)=-x^-1$$
I have found that $$u_2=fracsinx^22int textcosec^2x dx=-fraccosx^22$$
Now I tried to replicate this using the substitution $u_2=u_1v(x)$.
After finding the first and second derivatives of this equation and substituting into $(1)$, I get
$$v''sinx^2+v'left(4xcosx^2-x^-1sinx^2right)=0$$
Letting $w=v'$,
$$fracdwdxsinx^2+wleft(4xcosx^2-x^-1sinx^2right)=0$$
I tried to simplify this using the integrating factor
$$textexpleft(int 4xcosx^2-x^-1sinx^2 dxright)$$ but could not compute it.
How can I solve this problem using my suggested substitution?
calculus differential-equations reduction-of-order-ode
asked Jul 29 at 6:44
Bell
565113
Suppose $u_1=sinx^2$ is a solution of
$$xu''-u'+4x^3u=0Rightarrow u''-x^-1u'+4x^2u=0 (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_1intfracdxu^2_1textexpleft(int p(x) dxright) , p(x)=-x^-1$$
I have found that $$u_2=fracsinx^22int textcosec^2x dx=-fraccosx^22$$
Now I tried to replicate this using the substitution $u_2=u_1v(x)$.
After finding the first and second derivatives of this equation and substituting into $(1)$, I get
$$v''sinx^2+v'left(4xcosx^2-x^-1sinx^2right)=0$$
Letting $w=v'$,
$$fracdwdxsinx^2+wleft(4xcosx^2-x^-1sinx^2right)=0$$
I tried to simplify this using the integrating factor
$$textexpleft(int 4xcosx^2-x^-1sinx^2 dxright)$$ but could not compute it.
How can I solve this problem using my suggested substitution?
calculus differential-equations reduction-of-order-ode
asked Jul 29 at 6:44
Bell
565113
asked Jul 29 at 6:44
Bell
565113
asked Jul 29 at 6:44
Bell
565113
asked Jul 29 at 6:44
Bell
565113
565113
add a comment |Â
add a comment |Â
2 Answers
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The ODE for $w$ is actually separable:
beginalign*
sin(x^2)fracdwdx + left(4xcos(x^2) - fracsin(x^2)xright)w & = 0 \
int frac1w, dw & = int frac1sin(x^2)left(fracsin(x^2)x - 4xcos(x^2)right), dx \
intfrac1w, dw & = intfrac1x - 4xcot(x^2), dx \
endalign*
Making a change of variable $s = x^2$ on the second integrand, we obtain
beginalign*
intfrac1w, dw & = int frac1x, dx - 2intcot(s), ds \
ln|w| & = ln|x| - 2ln|sin(x^2)| + C
endalign*
You may then exponentiate each side and get
$$ |w| = e^Cfracxsin^2(x^2) = fracxsin^2(x^2). $$
answered Jul 29 at 7:28
Chee Han
2,4601616
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
 |Â
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up vote
1
down vote
Here is a solution without the reduction method. 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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-frac1X,D+4X^2=left(D+2texti,X-frac1Xright),(D-2texti,X),,$$
where $texti:=sqrt-1$.
If $u$ is a solution to $displaystyle left(D^2-frac1X,D+4X^2right),u=0$, we set $v:=(D-2texti,X),u$, so that
$$left(D+2texti,X-frac1Xright),v=0,,text or D,left(fracexp(texti,X^2)X,vright)=0,.$$
Thus, $v(x)=-4texti,a,x,expleft(-textix^2right)$ for some constant $a$. Now, $(D-2texti,X),u=v$ gives
$$D,big(exp(-texti,X^2),ubig)=exp(-texti,X^2),v,,text whence u(x)=exp(+texti,x^2),int,exp(-texti,x^2),v(x),textdx,.$$
Consequently, for some constant $b$, we get
$$u(x)=exp(+texti,x^2),int,(-4texti,a,x),expleft(-2texti,x^2right),textdx=exp(+texti,x^2),big(a,exp(-2texti,x^2)+bbig),.$$
That is,
$$u(x)=a,exp(-texti,x^2)+b,exp(+texti,x^2)=(a+b),cos(x^2)-texti,(a-b),sin(x^2),.$$
answered Aug 2 at 13:58
Batominovski
22.9k22777
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The ODE for $w$ is actually separable:
beginalign*
sin(x^2)fracdwdx + left(4xcos(x^2) - fracsin(x^2)xright)w & = 0 \
int frac1w, dw & = int frac1sin(x^2)left(fracsin(x^2)x - 4xcos(x^2)right), dx \
intfrac1w, dw & = intfrac1x - 4xcot(x^2), dx \
endalign*
Making a change of variable $s = x^2$ on the second integrand, we obtain
beginalign*
intfrac1w, dw & = int frac1x, dx - 2intcot(s), ds \
ln|w| & = ln|x| - 2ln|sin(x^2)| + C
endalign*
You may then exponentiate each side and get
$$ |w| = e^Cfracxsin^2(x^2) = fracxsin^2(x^2). $$
answered Jul 29 at 7:28
Chee Han
2,4601616
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
 |Â
show 3 more comments
up vote
2
down vote
accepted
The ODE for $w$ is actually separable:
beginalign*
sin(x^2)fracdwdx + left(4xcos(x^2) - fracsin(x^2)xright)w & = 0 \
int frac1w, dw & = int frac1sin(x^2)left(fracsin(x^2)x - 4xcos(x^2)right), dx \
intfrac1w, dw & = intfrac1x - 4xcot(x^2), dx \
endalign*
Making a change of variable $s = x^2$ on the second integrand, we obtain
beginalign*
intfrac1w, dw & = int frac1x, dx - 2intcot(s), ds \
ln|w| & = ln|x| - 2ln|sin(x^2)| + C
endalign*
You may then exponentiate each side and get
$$ |w| = e^Cfracxsin^2(x^2) = fracxsin^2(x^2). $$
answered Jul 29 at 7:28
Chee Han
2,4601616
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
 |Â
show 3 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The ODE for $w$ is actually separable:
beginalign*
sin(x^2)fracdwdx + left(4xcos(x^2) - fracsin(x^2)xright)w & = 0 \
int frac1w, dw & = int frac1sin(x^2)left(fracsin(x^2)x - 4xcos(x^2)right), dx \
intfrac1w, dw & = intfrac1x - 4xcot(x^2), dx \
endalign*
Making a change of variable $s = x^2$ on the second integrand, we obtain
beginalign*
intfrac1w, dw & = int frac1x, dx - 2intcot(s), ds \
ln|w| & = ln|x| - 2ln|sin(x^2)| + C
endalign*
You may then exponentiate each side and get
$$ |w| = e^Cfracxsin^2(x^2) = fracxsin^2(x^2). $$
answered Jul 29 at 7:28
Chee Han
2,4601616
The ODE for $w$ is actually separable:
beginalign*
sin(x^2)fracdwdx + left(4xcos(x^2) - fracsin(x^2)xright)w & = 0 \
int frac1w, dw & = int frac1sin(x^2)left(fracsin(x^2)x - 4xcos(x^2)right), dx \
intfrac1w, dw & = intfrac1x - 4xcot(x^2), dx \
endalign*
Making a change of variable $s = x^2$ on the second integrand, we obtain
beginalign*
intfrac1w, dw & = int frac1x, dx - 2intcot(s), ds \
ln|w| & = ln|x| - 2ln|sin(x^2)| + C
endalign*
You may then exponentiate each side and get
$$ |w| = e^Cfracxsin^2(x^2) = fracxsin^2(x^2). $$
answered Jul 29 at 7:28
Chee Han
2,4601616
answered Jul 29 at 7:28
Chee Han
2,4601616
answered Jul 29 at 7:28
Chee Han
2,4601616
answered Jul 29 at 7:28
Chee Han
2,4601616
2,4601616
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
 |Â
show 3 more comments
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
Awesome, I overlooked that it was separable. Thank you! After computing $w$, I substituted $w=v'$ and and obtained $$v=-fracAcotx^22+C_1$$ Substituting this back into $u_2=sinx^2v$, I get that $$u_2=frac-A2cosx^2+sinx^2C_1$$ Is this correct? It does satisfy $(1)$.
– Bell
Jul 29 at 8:29
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
$u_2=-fracA2cosx^2+sinx^2C_1$ is indeed a solution to $(1)$ but it is not linearly independent to $u_1=sinx^2$ in the case where $A=0$ and $C_1=1$. It is however a linearly independent solution $forall AinmathbbR$ (except $A=0$). If we take $C_1=0$, we are left with $u_2=-fracA2cosx^2$ which satisfies $(1)$ and is linearly independent to $u_1$. Why does setting $C_1=0$ work?
– Bell
Jul 29 at 9:05
1
1
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
@Bell because the given ODE is linear, i.e. if $u_1$ and $u_2$ are solutions then any linear combination of $u_1$ and $u_2$ is also a solution. You could then just "absorb" the $sin(x^2)$ term from $u_2$ into $u_1$.
– Chee Han
Jul 29 at 19:40
1
1
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
@Bell I should probably be more precise: I meant to say linear homogeneous ODE. Try to think of it in terms of differential operator. Suppose the ODE can be written as $Lu = 0$, where $L$ is your linear differential operator. If $u_1, u_2$ are two solutions of $Lu = 0$, then for any scalar $C, D$ we have that $L(Cu_1 + Du_2) = CL(u_1) + DL(u_2) = 0$, i.e. $Cu_1 + Du_2$ is also a solution of $Lu = 0$.
– Chee Han
Jul 30 at 21:38
1
1
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
@Bell Yes that is correct.
– Chee Han
Jul 31 at 2:21
 |Â
show 3 more comments
up vote
1
down vote
Here is a solution without the reduction method. 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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-frac1X,D+4X^2=left(D+2texti,X-frac1Xright),(D-2texti,X),,$$
where $texti:=sqrt-1$.
If $u$ is a solution to $displaystyle left(D^2-frac1X,D+4X^2right),u=0$, we set $v:=(D-2texti,X),u$, so that
$$left(D+2texti,X-frac1Xright),v=0,,text or D,left(fracexp(texti,X^2)X,vright)=0,.$$
Thus, $v(x)=-4texti,a,x,expleft(-textix^2right)$ for some constant $a$. Now, $(D-2texti,X),u=v$ gives
$$D,big(exp(-texti,X^2),ubig)=exp(-texti,X^2),v,,text whence u(x)=exp(+texti,x^2),int,exp(-texti,x^2),v(x),textdx,.$$
Consequently, for some constant $b$, we get
$$u(x)=exp(+texti,x^2),int,(-4texti,a,x),expleft(-2texti,x^2right),textdx=exp(+texti,x^2),big(a,exp(-2texti,x^2)+bbig),.$$
That is,
$$u(x)=a,exp(-texti,x^2)+b,exp(+texti,x^2)=(a+b),cos(x^2)-texti,(a-b),sin(x^2),.$$
answered Aug 2 at 13:58
Batominovski
22.9k22777
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up vote
1
down vote
Here is a solution without the reduction method. 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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-frac1X,D+4X^2=left(D+2texti,X-frac1Xright),(D-2texti,X),,$$
where $texti:=sqrt-1$.
If $u$ is a solution to $displaystyle left(D^2-frac1X,D+4X^2right),u=0$, we set $v:=(D-2texti,X),u$, so that
$$left(D+2texti,X-frac1Xright),v=0,,text or D,left(fracexp(texti,X^2)X,vright)=0,.$$
Thus, $v(x)=-4texti,a,x,expleft(-textix^2right)$ for some constant $a$. Now, $(D-2texti,X),u=v$ gives
$$D,big(exp(-texti,X^2),ubig)=exp(-texti,X^2),v,,text whence u(x)=exp(+texti,x^2),int,exp(-texti,x^2),v(x),textdx,.$$
Consequently, for some constant $b$, we get
$$u(x)=exp(+texti,x^2),int,(-4texti,a,x),expleft(-2texti,x^2right),textdx=exp(+texti,x^2),big(a,exp(-2texti,x^2)+bbig),.$$
That is,
$$u(x)=a,exp(-texti,x^2)+b,exp(+texti,x^2)=(a+b),cos(x^2)-texti,(a-b),sin(x^2),.$$
answered Aug 2 at 13:58
Batominovski
22.9k22777
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up vote
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down vote
Here is a solution without the reduction method. 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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-frac1X,D+4X^2=left(D+2texti,X-frac1Xright),(D-2texti,X),,$$
where $texti:=sqrt-1$.
If $u$ is a solution to $displaystyle left(D^2-frac1X,D+4X^2right),u=0$, we set $v:=(D-2texti,X),u$, so that
$$left(D+2texti,X-frac1Xright),v=0,,text or D,left(fracexp(texti,X^2)X,vright)=0,.$$
Thus, $v(x)=-4texti,a,x,expleft(-textix^2right)$ for some constant $a$. Now, $(D-2texti,X),u=v$ gives
$$D,big(exp(-texti,X^2),ubig)=exp(-texti,X^2),v,,text whence u(x)=exp(+texti,x^2),int,exp(-texti,x^2),v(x),textdx,.$$
Consequently, for some constant $b$, we get
$$u(x)=exp(+texti,x^2),int,(-4texti,a,x),expleft(-2texti,x^2right),textdx=exp(+texti,x^2),big(a,exp(-2texti,x^2)+bbig),.$$
That is,
$$u(x)=a,exp(-texti,x^2)+b,exp(+texti,x^2)=(a+b),cos(x^2)-texti,(a-b),sin(x^2),.$$
answered Aug 2 at 13:58
Batominovski
22.9k22777
Here is a solution without the reduction method. 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 $big(phi(X),hbig)(x):=phi(x),h(x)$. Observe that
$$D^2-frac1X,D+4X^2=left(D+2texti,X-frac1Xright),(D-2texti,X),,$$
where $texti:=sqrt-1$.
If $u$ is a solution to $displaystyle left(D^2-frac1X,D+4X^2right),u=0$, we set $v:=(D-2texti,X),u$, so that
$$left(D+2texti,X-frac1Xright),v=0,,text or D,left(fracexp(texti,X^2)X,vright)=0,.$$
Thus, $v(x)=-4texti,a,x,expleft(-textix^2right)$ for some constant $a$. Now, $(D-2texti,X),u=v$ gives
$$D,big(exp(-texti,X^2),ubig)=exp(-texti,X^2),v,,text whence u(x)=exp(+texti,x^2),int,exp(-texti,x^2),v(x),textdx,.$$
Consequently, for some constant $b$, we get
$$u(x)=exp(+texti,x^2),int,(-4texti,a,x),expleft(-2texti,x^2right),textdx=exp(+texti,x^2),big(a,exp(-2texti,x^2)+bbig),.$$
That is,
$$u(x)=a,exp(-texti,x^2)+b,exp(+texti,x^2)=(a+b),cos(x^2)-texti,(a-b),sin(x^2),.$$
answered Aug 2 at 13:58
Batominovski
22.9k22777
answered Aug 2 at 13:58
Batominovski
22.9k22777
answered Aug 2 at 13:58
Batominovski
22.9k22777
answered Aug 2 at 13:58
Batominovski
22.9k22777
22.9k22777
add a comment |Â
add a comment |Â
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