Use Laplace transform to solve $xyâ€Â+(1-x)y’+my=0$ [closed]
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Use Laplace transform to solve $xy''+(1-x)y'+my=0$.
(a) $y=displaystylefrace^tk! fracd^kdt^k(t^-ke^-t)$
(b) $y=displaystylefrace^tk fracd^kdt^k(t^k e^-t)$
(c) $y=displaystylefrace^tk! fracd^kdt^k(t^k e^-t)$
differential-equations laplace-transform
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
asked Aug 20 at 5:58
Huang
1
closed as off-topic by Chickenmancer, uniquesolution, Kavi Rama Murthy, Nosrati, Batominovski Aug 20 at 12:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – uniquesolution, Kavi Rama Murthy, Nosrati
add a comment |Â
up vote
-1
down vote
favorite
Use Laplace transform to solve $xy''+(1-x)y'+my=0$.
(a) $y=displaystylefrace^tk! fracd^kdt^k(t^-ke^-t)$
(b) $y=displaystylefrace^tk fracd^kdt^k(t^k e^-t)$
(c) $y=displaystylefrace^tk! fracd^kdt^k(t^k e^-t)$
differential-equations laplace-transform
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
asked Aug 20 at 5:58
Huang
1
closed as off-topic by Chickenmancer, uniquesolution, Kavi Rama Murthy, Nosrati, Batominovski Aug 20 at 12:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – uniquesolution, Kavi Rama Murthy, Nosrati
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Use Laplace transform to solve $xy''+(1-x)y'+my=0$.
(a) $y=displaystylefrace^tk! fracd^kdt^k(t^-ke^-t)$
(b) $y=displaystylefrace^tk fracd^kdt^k(t^k e^-t)$
(c) $y=displaystylefrace^tk! fracd^kdt^k(t^k e^-t)$
differential-equations laplace-transform
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
asked Aug 20 at 5:58
Huang
1
Use Laplace transform to solve $xy''+(1-x)y'+my=0$.
(a) $y=displaystylefrace^tk! fracd^kdt^k(t^-ke^-t)$
(b) $y=displaystylefrace^tk fracd^kdt^k(t^k e^-t)$
(c) $y=displaystylefrace^tk! fracd^kdt^k(t^k e^-t)$
differential-equations laplace-transform
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
asked Aug 20 at 5:58
Huang
1
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
edited Aug 20 at 12:44
Taroccoesbrocco
3,72651433
3,72651433
asked Aug 20 at 5:58
Huang
1
asked Aug 20 at 5:58
Huang
1
asked Aug 20 at 5:58
Huang
1
1
closed as off-topic by Chickenmancer, uniquesolution, Kavi Rama Murthy, Nosrati, Batominovski Aug 20 at 12:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – uniquesolution, Kavi Rama Murthy, Nosrati
closed as off-topic by Chickenmancer, uniquesolution, Kavi Rama Murthy, Nosrati, Batominovski Aug 20 at 12:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – uniquesolution, Kavi Rama Murthy, Nosrati
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44
add a comment |Â
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44
add a comment |Â
1 Answer
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Hint.
$$
mathcalLleft(x f(x)right) = -fracddsmathcalLleft(f(x)right)
$$
then
$$
-fracddsmathcalLleft(ddot yright)+fracddsmathcalLleft(dot yright)+mathcalLleft(dot yright)+mmathcalLleft(yright)=0
$$
with
$$
mathcalLleft(ddot yright) = s^2Y(s)-dot y(0)-s y(0)\
mathcalLleft(dot yright) = sY(s)-y(0)\
$$
and then solve the $Y(s)$ DE.
NOTE
After deriving we get at
$$
Y'(s)(s-s^2) + (m+1-s)Y(s) = 0
$$
with solution
$$
Y(s) = C_0 frac(1-s)^ms^m+1
$$
answered Aug 20 at 8:37
Cesareo
5,9052412
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint.
$$
mathcalLleft(x f(x)right) = -fracddsmathcalLleft(f(x)right)
$$
then
$$
-fracddsmathcalLleft(ddot yright)+fracddsmathcalLleft(dot yright)+mathcalLleft(dot yright)+mmathcalLleft(yright)=0
$$
with
$$
mathcalLleft(ddot yright) = s^2Y(s)-dot y(0)-s y(0)\
mathcalLleft(dot yright) = sY(s)-y(0)\
$$
and then solve the $Y(s)$ DE.
NOTE
After deriving we get at
$$
Y'(s)(s-s^2) + (m+1-s)Y(s) = 0
$$
with solution
$$
Y(s) = C_0 frac(1-s)^ms^m+1
$$
answered Aug 20 at 8:37
Cesareo
5,9052412
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
add a comment |Â
up vote
1
down vote
Hint.
$$
mathcalLleft(x f(x)right) = -fracddsmathcalLleft(f(x)right)
$$
then
$$
-fracddsmathcalLleft(ddot yright)+fracddsmathcalLleft(dot yright)+mathcalLleft(dot yright)+mmathcalLleft(yright)=0
$$
with
$$
mathcalLleft(ddot yright) = s^2Y(s)-dot y(0)-s y(0)\
mathcalLleft(dot yright) = sY(s)-y(0)\
$$
and then solve the $Y(s)$ DE.
NOTE
After deriving we get at
$$
Y'(s)(s-s^2) + (m+1-s)Y(s) = 0
$$
with solution
$$
Y(s) = C_0 frac(1-s)^ms^m+1
$$
answered Aug 20 at 8:37
Cesareo
5,9052412
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint.
$$
mathcalLleft(x f(x)right) = -fracddsmathcalLleft(f(x)right)
$$
then
$$
-fracddsmathcalLleft(ddot yright)+fracddsmathcalLleft(dot yright)+mathcalLleft(dot yright)+mmathcalLleft(yright)=0
$$
with
$$
mathcalLleft(ddot yright) = s^2Y(s)-dot y(0)-s y(0)\
mathcalLleft(dot yright) = sY(s)-y(0)\
$$
and then solve the $Y(s)$ DE.
NOTE
After deriving we get at
$$
Y'(s)(s-s^2) + (m+1-s)Y(s) = 0
$$
with solution
$$
Y(s) = C_0 frac(1-s)^ms^m+1
$$
answered Aug 20 at 8:37
Cesareo
5,9052412
Hint.
$$
mathcalLleft(x f(x)right) = -fracddsmathcalLleft(f(x)right)
$$
then
$$
-fracddsmathcalLleft(ddot yright)+fracddsmathcalLleft(dot yright)+mathcalLleft(dot yright)+mmathcalLleft(yright)=0
$$
with
$$
mathcalLleft(ddot yright) = s^2Y(s)-dot y(0)-s y(0)\
mathcalLleft(dot yright) = sY(s)-y(0)\
$$
and then solve the $Y(s)$ DE.
NOTE
After deriving we get at
$$
Y'(s)(s-s^2) + (m+1-s)Y(s) = 0
$$
with solution
$$
Y(s) = C_0 frac(1-s)^ms^m+1
$$
answered Aug 20 at 8:37
Cesareo
5,9052412
edited Aug 20 at 11:42
answered Aug 20 at 8:37
Cesareo
5,9052412
answered Aug 20 at 8:37
Cesareo
5,9052412
answered Aug 20 at 8:37
Cesareo
5,9052412
5,9052412
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
add a comment |Â
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
So is it a, b, or c!!!! That's the real question.
– user14717
Aug 20 at 10:40
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
@user14717 This is your attribution. See the attached note
– Cesareo
Aug 20 at 11:39
1
1
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
@user14717 Cesareo gave a good hint, and that should have been sufficient for you to work it out. If you had tried, then you would have seen that $$y(x)=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$$ is a solution.
– Batominovski
Aug 20 at 12:39
1
1
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
Furthermore, if $displaystyle f_m(x):=fracexp(x)m!,fractextd^mtextdx^m,big(x^m,exp(-x)big)$, then all solutions to the differential equation $$x,y''(x)+(1-x),y'(x)+m,y'(x)=0,, text where minmathbbZ_geq 0,,$$ takes the form $y(x)=A,f_m(x)+B,g_m(x)$, where $A$ and $B$ are constants, and $$g_m(x):=f_m(x),int_1^x,fracexp(t)t,big(f_m(t)big)^2,textdt,.$$
– Batominovski
Aug 20 at 12:56
It was a joke, people.
– user14717
Aug 21 at 1:26
It was a joke, people.
– user14717
Aug 21 at 1:26
add a comment |Â
Welcome to math.stackexchange. Please be aware that this is not a homework working service. Please explain your thoughts on the problem and what difficulties you are having finding the solution and perhaps someone will be willing to help you sort it out.
– John Wayland Bales
Aug 20 at 6:27
What is $k$? Did you mean $m$? $k$ can't for sure be arbitrary.
– Batominovski
Aug 20 at 8:44