How can the following coupled system of ODE be solved?

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The following system is a coupled ODEs that are dependant on each other.

$$ Y''(x)+A_1 (Y(x) - Z(x)) = A_2 Y(x) - H(x -A_3)$$
$$ Z''(x)+B_1 (Z(x) - Y(x)) = B_2 Z(x) - H(x -B_3)$$

Where:

H(x) is the Unit Step Function

differential-equations

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edited Aug 6 at 1:27

asked Aug 6 at 1:04

Refaat Galal

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The following system is a coupled ODEs that are dependant on each other.

$$ Y''(x)+A_1 (Y(x) - Z(x)) = A_2 Y(x) - H(x -A_3)$$
$$ Z''(x)+B_1 (Z(x) - Y(x)) = B_2 Z(x) - H(x -B_3)$$

Where:

H(x) is the Unit Step Function

differential-equations

share|cite|improve this question

edited Aug 6 at 1:27

asked Aug 6 at 1:04

Refaat Galal

93

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The following system is a coupled ODEs that are dependant on each other.

$$ Y''(x)+A_1 (Y(x) - Z(x)) = A_2 Y(x) - H(x -A_3)$$
$$ Z''(x)+B_1 (Z(x) - Y(x)) = B_2 Z(x) - H(x -B_3)$$

Where:

H(x) is the Unit Step Function

differential-equations

share|cite|improve this question

edited Aug 6 at 1:27

asked Aug 6 at 1:04

Refaat Galal

93

The following system is a coupled ODEs that are dependant on each other.

$$ Y''(x)+A_1 (Y(x) - Z(x)) = A_2 Y(x) - H(x -A_3)$$
$$ Z''(x)+B_1 (Z(x) - Y(x)) = B_2 Z(x) - H(x -B_3)$$

Where:

H(x) is the Unit Step Function

differential-equations

share|cite|improve this question

edited Aug 6 at 1:27

asked Aug 6 at 1:04

Refaat Galal

93

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 1:27

edited Aug 6 at 1:27

edited Aug 6 at 1:27

asked Aug 6 at 1:04

Refaat Galal

93

asked Aug 6 at 1:04

Refaat Galal

93

asked Aug 6 at 1:04

Refaat Galal

93

93

add a comment | 

add a comment | 

1 Answer
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Let $begincases
y_1(x) = Y(x)\
y_2(x) = y_1'(x)\
z_1(x) = Z(x)\
z_2(x) = z_1'(x)
endcases$, you may rewrite as,

$begincases
y_1(x)' = y_2(x)\
y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\
z_1(x)' = z_2(x)\
z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\
h_1(x) = H(x - A_3)\
h_2(x) = H(x - B_3)
endcases$

and

$
left(
beginarrayc
y_1(x)'\
y_2(x)'\
z_1(x)'\
z_2(x)'\
h_1(x)\
h_2(x)
endarray
right)
=
left(
beginarraycccccc
0 & 1 & 0 & 0 & 0 & 0\
(A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \
0 & 0 & 0 & 1 & 0 & 0\
B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\
0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 1
endarray
right)
left(
beginarrayc
Y(x)\
Y'(x)\
Z(x)\
Z'(x)\
H(x - A_3)\
H(x - B_3)
endarray
right)
$

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answered Aug 6 at 7:07

GinoCHJ

794

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Let $begincases
y_1(x) = Y(x)\
y_2(x) = y_1'(x)\
z_1(x) = Z(x)\
z_2(x) = z_1'(x)
endcases$, you may rewrite as,

$begincases
y_1(x)' = y_2(x)\
y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\
z_1(x)' = z_2(x)\
z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\
h_1(x) = H(x - A_3)\
h_2(x) = H(x - B_3)
endcases$

and

$
left(
beginarrayc
y_1(x)'\
y_2(x)'\
z_1(x)'\
z_2(x)'\
h_1(x)\
h_2(x)
endarray
right)
=
left(
beginarraycccccc
0 & 1 & 0 & 0 & 0 & 0\
(A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \
0 & 0 & 0 & 1 & 0 & 0\
B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\
0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 1
endarray
right)
left(
beginarrayc
Y(x)\
Y'(x)\
Z(x)\
Z'(x)\
H(x - A_3)\
H(x - B_3)
endarray
right)
$

share|cite|improve this answer

answered Aug 6 at 7:07

GinoCHJ

794

add a comment | 

up vote
0
down vote

Let $begincases
y_1(x) = Y(x)\
y_2(x) = y_1'(x)\
z_1(x) = Z(x)\
z_2(x) = z_1'(x)
endcases$, you may rewrite as,

$begincases
y_1(x)' = y_2(x)\
y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\
z_1(x)' = z_2(x)\
z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\
h_1(x) = H(x - A_3)\
h_2(x) = H(x - B_3)
endcases$

and

$
left(
beginarrayc
y_1(x)'\
y_2(x)'\
z_1(x)'\
z_2(x)'\
h_1(x)\
h_2(x)
endarray
right)
=
left(
beginarraycccccc
0 & 1 & 0 & 0 & 0 & 0\
(A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \
0 & 0 & 0 & 1 & 0 & 0\
B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\
0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 1
endarray
right)
left(
beginarrayc
Y(x)\
Y'(x)\
Z(x)\
Z'(x)\
H(x - A_3)\
H(x - B_3)
endarray
right)
$

share|cite|improve this answer

answered Aug 6 at 7:07

GinoCHJ

794

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $begincases
y_1(x) = Y(x)\
y_2(x) = y_1'(x)\
z_1(x) = Z(x)\
z_2(x) = z_1'(x)
endcases$, you may rewrite as,

$begincases
y_1(x)' = y_2(x)\
y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\
z_1(x)' = z_2(x)\
z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\
h_1(x) = H(x - A_3)\
h_2(x) = H(x - B_3)
endcases$

and

$
left(
beginarrayc
y_1(x)'\
y_2(x)'\
z_1(x)'\
z_2(x)'\
h_1(x)\
h_2(x)
endarray
right)
=
left(
beginarraycccccc
0 & 1 & 0 & 0 & 0 & 0\
(A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \
0 & 0 & 0 & 1 & 0 & 0\
B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\
0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 1
endarray
right)
left(
beginarrayc
Y(x)\
Y'(x)\
Z(x)\
Z'(x)\
H(x - A_3)\
H(x - B_3)
endarray
right)
$

share|cite|improve this answer

answered Aug 6 at 7:07

GinoCHJ

794

Let $begincases
y_1(x) = Y(x)\
y_2(x) = y_1'(x)\
z_1(x) = Z(x)\
z_2(x) = z_1'(x)
endcases$, you may rewrite as,

$begincases
y_1(x)' = y_2(x)\
y_2(x)' = (A_2 - A_1)y_1(x) + A_1 z_1(x) - h_1(x)\
z_1(x)' = z_2(x)\
z_2(x)' = (B_2-B_1)z_1(x) + B_1 y_1(x)) + z_1(x) - h_2(x)\
h_1(x) = H(x - A_3)\
h_2(x) = H(x - B_3)
endcases$

and

$
left(
beginarrayc
y_1(x)'\
y_2(x)'\
z_1(x)'\
z_2(x)'\
h_1(x)\
h_2(x)
endarray
right)
=
left(
beginarraycccccc
0 & 1 & 0 & 0 & 0 & 0\
(A_2 - A_1)& 0 & A_1 & 0 & -1 & 0 \
0 & 0 & 0 & 1 & 0 & 0\
B_1 & 0 & (B_2-B_1) & 0 & 0 &-1\
0 & 0 & 0 & 0 & 1 & 0\
0 & 0 & 0 & 0 & 0 & 1
endarray
right)
left(
beginarrayc
Y(x)\
Y'(x)\
Z(x)\
Z'(x)\
H(x - A_3)\
H(x - B_3)
endarray
right)
$

share|cite|improve this answer

answered Aug 6 at 7:07

GinoCHJ

794

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 7:07

GinoCHJ

794

answered Aug 6 at 7:07

GinoCHJ

794

answered Aug 6 at 7:07

GinoCHJ

794

794

add a comment | 

add a comment | 

 

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