Transform a non-linear differential equation into a linear equation
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Following the work of Yaoji Lu - 1967 (here's a link to the full paper) I got stuck at the step when the author transform a non-linear differential equation into a linear equation (eq. 3.9 pag. 19).
More precisely he starts from this equation:
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Then define $Z=Y^{1-frac{1}{b}}$ and eventually get:
$$frac{dZ}{dX}+frac{1-b}{b}cdot frac{1}{X}cdot Z=alpha frac{1-b}{b}cdot X^{-frac{c}{b}-1}$$
Is there anyone who can help me with this?
linear-algebra differential-equations transformation economics nonlinear-analysis
asked 2 days ago
Alessandro
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Following the work of Yaoji Lu - 1967 (here's a link to the full paper) I got stuck at the step when the author transform a non-linear differential equation into a linear equation (eq. 3.9 pag. 19).
More precisely he starts from this equation:
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Then define $Z=Y^{1-frac{1}{b}}$ and eventually get:
$$frac{dZ}{dX}+frac{1-b}{b}cdot frac{1}{X}cdot Z=alpha frac{1-b}{b}cdot X^{-frac{c}{b}-1}$$
Is there anyone who can help me with this?
linear-algebra differential-equations transformation economics nonlinear-analysis
asked 2 days ago
Alessandro
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It's Bernouilli 's equation
– Isham
2 days ago
add a comment |
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up vote
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down vote
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Following the work of Yaoji Lu - 1967 (here's a link to the full paper) I got stuck at the step when the author transform a non-linear differential equation into a linear equation (eq. 3.9 pag. 19).
More precisely he starts from this equation:
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Then define $Z=Y^{1-frac{1}{b}}$ and eventually get:
$$frac{dZ}{dX}+frac{1-b}{b}cdot frac{1}{X}cdot Z=alpha frac{1-b}{b}cdot X^{-frac{c}{b}-1}$$
Is there anyone who can help me with this?
linear-algebra differential-equations transformation economics nonlinear-analysis
asked 2 days ago
Alessandro
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Following the work of Yaoji Lu - 1967 (here's a link to the full paper) I got stuck at the step when the author transform a non-linear differential equation into a linear equation (eq. 3.9 pag. 19).
More precisely he starts from this equation:
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Then define $Z=Y^{1-frac{1}{b}}$ and eventually get:
$$frac{dZ}{dX}+frac{1-b}{b}cdot frac{1}{X}cdot Z=alpha frac{1-b}{b}cdot X^{-frac{c}{b}-1}$$
Is there anyone who can help me with this?
linear-algebra differential-equations transformation economics nonlinear-analysis
linear-algebra differential-equations transformation economics nonlinear-analysis
asked 2 days ago
Alessandro
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Alessandro
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Alessandro
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Alessandro
11
asked 2 days ago
Alessandro
11
11
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alessandro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It's Bernouilli 's equation
– Isham
2 days ago
add a comment |
It's Bernouilli 's equation
– Isham
2 days ago
It's Bernouilli 's equation
– Isham
2 days ago
It's Bernouilli 's equation
– Isham
2 days ago
add a comment |
1 Answer
1
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0
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It's Bernouilli's equation
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Divide by $y^{1/b}$
$$Y^{-frac{1}{b}}Y'=frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} $$
$$(1-frac 1b )Y^{-frac{1}{b}}Y'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
$$(Y^{1-frac{1}{b}})'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
Substitute $Z=Y^{1-frac 1b}$
$$Z'=(1-frac 1b)(frac Z{X}-alpha cdot X^{-frac{c}{b}-1} )$$
answered 2 days ago
Isham
12.7k3929
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It's Bernouilli's equation
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Divide by $y^{1/b}$
$$Y^{-frac{1}{b}}Y'=frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} $$
$$(1-frac 1b )Y^{-frac{1}{b}}Y'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
$$(Y^{1-frac{1}{b}})'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
Substitute $Z=Y^{1-frac 1b}$
$$Z'=(1-frac 1b)(frac Z{X}-alpha cdot X^{-frac{c}{b}-1} )$$
answered 2 days ago
Isham
12.7k3929
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
add a comment |
up vote
0
down vote
It's Bernouilli's equation
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Divide by $y^{1/b}$
$$Y^{-frac{1}{b}}Y'=frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} $$
$$(1-frac 1b )Y^{-frac{1}{b}}Y'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
$$(Y^{1-frac{1}{b}})'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
Substitute $Z=Y^{1-frac 1b}$
$$Z'=(1-frac 1b)(frac Z{X}-alpha cdot X^{-frac{c}{b}-1} )$$
answered 2 days ago
Isham
12.7k3929
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
It's Bernouilli's equation
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Divide by $y^{1/b}$
$$Y^{-frac{1}{b}}Y'=frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} $$
$$(1-frac 1b )Y^{-frac{1}{b}}Y'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
$$(Y^{1-frac{1}{b}})'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
Substitute $Z=Y^{1-frac 1b}$
$$Z'=(1-frac 1b)(frac Z{X}-alpha cdot X^{-frac{c}{b}-1} )$$
answered 2 days ago
Isham
12.7k3929
It's Bernouilli's equation
$$frac{dY}{dX}=frac{Y}{X}-alpha cdot X^{-frac{c}{b}-1}cdot Y^{frac{1}{b}}$$
Divide by $y^{1/b}$
$$Y^{-frac{1}{b}}Y'=frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} $$
$$(1-frac 1b )Y^{-frac{1}{b}}Y'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
$$(Y^{1-frac{1}{b}})'=(1-frac 1b)(frac{Y^{1-frac 1b}}{X}-alpha cdot X^{-frac{c}{b}-1} )$$
Substitute $Z=Y^{1-frac 1b}$
$$Z'=(1-frac 1b)(frac Z{X}-alpha cdot X^{-frac{c}{b}-1} )$$
answered 2 days ago
Isham
12.7k3929
answered 2 days ago
Isham
12.7k3929
answered 2 days ago
Isham
12.7k3929
answered 2 days ago
Isham
12.7k3929
12.7k3929
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
add a comment |
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
Thank you so much! ;)
– Alessandro
yesterday
Thank you so much! ;)
– Alessandro
yesterday
toy're welcome @Alessandro
– Isham
yesterday
toy're welcome @Alessandro
– Isham
yesterday
add a comment |
Alessandro is a new contributor. Be nice, and check out our Code of Conduct.
Alessandro is a new contributor. Be nice, and check out our Code of Conduct.
Alessandro is a new contributor. Be nice, and check out our Code of Conduct.
Alessandro is a new contributor. Be nice, and check out our Code of Conduct.
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It's Bernouilli 's equation
– Isham
2 days ago