Solve the given differential equation $(D^2+1)^2y= 24 x cos x$
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$$(D^2+1)^2y= 24 x cos x$$
given that $y=Dy=D^2y=0$ and $D^3 y=12$ when $x=0$
A part of the solution I got but finding difficulty in the other part.
Let it be y= y1 +y2
$y1=(A+Bx)sin(x)+(C+Dx)cos(x)$
For Integral part i was struck at
$$y2=(1/(D^2+1)^2)24x cosx $$
$$y2=-12i (1/(D^2+1))[(1/D-i)-(1/D+i)]xe^ix$$
The answer is $y=3x^2sinx-x^3cosx$
differential-equations
asked Sep 5 at 6:53
Bharath Teja
12
add a comment |Â
up vote
-1
down vote
favorite
$$(D^2+1)^2y= 24 x cos x$$
given that $y=Dy=D^2y=0$ and $D^3 y=12$ when $x=0$
A part of the solution I got but finding difficulty in the other part.
Let it be y= y1 +y2
$y1=(A+Bx)sin(x)+(C+Dx)cos(x)$
For Integral part i was struck at
$$y2=(1/(D^2+1)^2)24x cosx $$
$$y2=-12i (1/(D^2+1))[(1/D-i)-(1/D+i)]xe^ix$$
The answer is $y=3x^2sinx-x^3cosx$
differential-equations
asked Sep 5 at 6:53
Bharath Teja
12
1
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$$(D^2+1)^2y= 24 x cos x$$
given that $y=Dy=D^2y=0$ and $D^3 y=12$ when $x=0$
A part of the solution I got but finding difficulty in the other part.
Let it be y= y1 +y2
$y1=(A+Bx)sin(x)+(C+Dx)cos(x)$
For Integral part i was struck at
$$y2=(1/(D^2+1)^2)24x cosx $$
$$y2=-12i (1/(D^2+1))[(1/D-i)-(1/D+i)]xe^ix$$
The answer is $y=3x^2sinx-x^3cosx$
differential-equations
asked Sep 5 at 6:53
Bharath Teja
12
$$(D^2+1)^2y= 24 x cos x$$
given that $y=Dy=D^2y=0$ and $D^3 y=12$ when $x=0$
A part of the solution I got but finding difficulty in the other part.
Let it be y= y1 +y2
$y1=(A+Bx)sin(x)+(C+Dx)cos(x)$
For Integral part i was struck at
$$y2=(1/(D^2+1)^2)24x cosx $$
$$y2=-12i (1/(D^2+1))[(1/D-i)-(1/D+i)]xe^ix$$
The answer is $y=3x^2sinx-x^3cosx$
differential-equations
differential-equations
asked Sep 5 at 6:53
Bharath Teja
12
asked Sep 5 at 6:53
Bharath Teja
12
edited Sep 6 at 5:34
asked Sep 5 at 6:53
Bharath Teja
12
asked Sep 5 at 6:53
Bharath Teja
12
asked Sep 5 at 6:53
Bharath Teja
12
12
1
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24
add a comment |Â
1
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24
1
1
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
It's a linear differential equation with constant coefficients. This way you immediately know that the homogeneous solutions will be exponential/trigonometric and if there are any duplicate roots of the characteristic equations (there are, as you can see the left side is already factorized), you will get polynomial*exponential/trigonometric, too. The particular solution also follows the familiar procedure with trial functions in this case.
However, this question looks like it's made for Laplace transform. Just substitute $D$ with $s$ on the left, find the laplace transform of the right hand side from the tables, and invert the transform (you will need the initial conditions in this step).
answered Sep 5 at 7:08
orion
12.1k11832
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
add a comment |Â
up vote
1
down vote
Since the question shows no try, so this is just a hint. Take $$y=(A+Bx)sin(x)+(C+Dx)cos(x)$$ and find the unknown coefficients.
answered Sep 5 at 7:35
mrs
58.3k750143
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It's a linear differential equation with constant coefficients. This way you immediately know that the homogeneous solutions will be exponential/trigonometric and if there are any duplicate roots of the characteristic equations (there are, as you can see the left side is already factorized), you will get polynomial*exponential/trigonometric, too. The particular solution also follows the familiar procedure with trial functions in this case.
However, this question looks like it's made for Laplace transform. Just substitute $D$ with $s$ on the left, find the laplace transform of the right hand side from the tables, and invert the transform (you will need the initial conditions in this step).
answered Sep 5 at 7:08
orion
12.1k11832
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
add a comment |Â
up vote
1
down vote
It's a linear differential equation with constant coefficients. This way you immediately know that the homogeneous solutions will be exponential/trigonometric and if there are any duplicate roots of the characteristic equations (there are, as you can see the left side is already factorized), you will get polynomial*exponential/trigonometric, too. The particular solution also follows the familiar procedure with trial functions in this case.
However, this question looks like it's made for Laplace transform. Just substitute $D$ with $s$ on the left, find the laplace transform of the right hand side from the tables, and invert the transform (you will need the initial conditions in this step).
answered Sep 5 at 7:08
orion
12.1k11832
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's a linear differential equation with constant coefficients. This way you immediately know that the homogeneous solutions will be exponential/trigonometric and if there are any duplicate roots of the characteristic equations (there are, as you can see the left side is already factorized), you will get polynomial*exponential/trigonometric, too. The particular solution also follows the familiar procedure with trial functions in this case.
However, this question looks like it's made for Laplace transform. Just substitute $D$ with $s$ on the left, find the laplace transform of the right hand side from the tables, and invert the transform (you will need the initial conditions in this step).
answered Sep 5 at 7:08
orion
12.1k11832
It's a linear differential equation with constant coefficients. This way you immediately know that the homogeneous solutions will be exponential/trigonometric and if there are any duplicate roots of the characteristic equations (there are, as you can see the left side is already factorized), you will get polynomial*exponential/trigonometric, too. The particular solution also follows the familiar procedure with trial functions in this case.
However, this question looks like it's made for Laplace transform. Just substitute $D$ with $s$ on the left, find the laplace transform of the right hand side from the tables, and invert the transform (you will need the initial conditions in this step).
answered Sep 5 at 7:08
orion
12.1k11832
answered Sep 5 at 7:08
orion
12.1k11832
answered Sep 5 at 7:08
orion
12.1k11832
answered Sep 5 at 7:08
orion
12.1k11832
12.1k11832
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
add a comment |Â
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
Thanks for your nice comment. +
– mrs
Sep 5 at 10:37
add a comment |Â
up vote
1
down vote
Since the question shows no try, so this is just a hint. Take $$y=(A+Bx)sin(x)+(C+Dx)cos(x)$$ and find the unknown coefficients.
answered Sep 5 at 7:35
mrs
58.3k750143
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
add a comment |Â
up vote
1
down vote
Since the question shows no try, so this is just a hint. Take $$y=(A+Bx)sin(x)+(C+Dx)cos(x)$$ and find the unknown coefficients.
answered Sep 5 at 7:35
mrs
58.3k750143
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since the question shows no try, so this is just a hint. Take $$y=(A+Bx)sin(x)+(C+Dx)cos(x)$$ and find the unknown coefficients.
answered Sep 5 at 7:35
mrs
58.3k750143
Since the question shows no try, so this is just a hint. Take $$y=(A+Bx)sin(x)+(C+Dx)cos(x)$$ and find the unknown coefficients.
answered Sep 5 at 7:35
mrs
58.3k750143
answered Sep 5 at 7:35
mrs
58.3k750143
answered Sep 5 at 7:35
mrs
58.3k750143
answered Sep 5 at 7:35
mrs
58.3k750143
58.3k750143
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
add a comment |Â
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
This is part of solution. We need to find Particular integral too.,
– Bharath Teja
Sep 5 at 8:18
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
You see that $x cos x$ and $xsin x$ are already in the homogeneous solution. The procedure is then to increase the polynomial order until you get a match. Because $sin$ and $cos$ are not independent in this sense, you always need to include both. Try $(ax^2+bx+c)cos x+(dx^2+fx+g)sin x$.
– orion
Sep 5 at 8:36
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
I edited the question please see it.
– Bharath Teja
Sep 6 at 5:35
add a comment |Â
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1
What did you try and how far did you get? We need to know where the problem is.
– orion
Sep 5 at 7:00
This problem was asked in UPSC CSE mathematics optional paper. Its an exam for Civil services in India. I am finding difficulty in particular integral.
– Bharath Teja
Sep 5 at 8:24