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How to transform the differential equation? [Solved!]

My question

Dear friends, I met a problem.
I don't know how to transform the equation (13) into (14). Please see the link in pages 422-423.

Let y* be the steady-state level of income per effective worker given by equation
(11), and let y(t) be the actual value at time t.

`\frac{d ln(y(t))}{dt} = \lambda(ln(y"*") - lny(t))\ \ \ ` (13)

` ln(y(t)) = (1 - e^{-\lambda t} )ln(y"*") + e^{-\lambda t}lny(0)\ \ \ ` (14)

where

`lambda = (n + g + delta) (1 - alpha - beta)`, and

`y(0)` is income per effective worker at some initial date.

Relevant page

https://eml.berkeley.edu/~dromer/papers/MRW_QJE1992.pdf

What I've done so far

Sorry I don't know how to solve it.
I just assume `\lambda` is constant.

X

Dear friends, I met a problem.
I don't know how to transform the equation (13) into (14). Please see the link in pages 422-423.

Let y* be the steady-state level of income per effective worker given by equation
(11), and let y(t) be the actual value at time t. 

`\frac{d ln(y(t))}{dt} = \lambda(ln(y"*") - lny(t))\ \ \ ` (13)

` ln(y(t)) = (1 - e^{-\lambda t} )ln(y"*") + e^{-\lambda t}lny(0)\ \ \ ` (14)

where

`lambda = (n + g + delta) (1 - alpha - beta)`, and

`y(0)` is income per effective worker at some initial date.
Relevant page

<a href="https://eml.berkeley.edu/~dromer/papers/MRW_QJE1992.pdf">https://eml.berkeley.edu/~dromer/papers/MRW_QJE1992.pdf</a>

What I've done so far

Sorry I don't know how to solve it.
I just assume `\lambda` is constant.

Re: How to transform the differential equation?

@zangyhui: I edited your question so it was easier for others to follow.

It indicates in the "where" expression that `lambda` is indeed a constant since each of the letters in the expression represent constants.

This page will help you to connect the dots: Linear DEs of Order 1

To help you get started, and to make the writing easier, I would suggest you make these substitutions:

`z = ln(y(t))`

So the left hand side of (13) is now: `(dz)/(dt).`

`Q = lambda ln(y"*")`

Can you write what `P` is? And can you go on from there?

X

@zangyhui: I edited your question so it was easier for others to follow.

It indicates in the "where" expression that `lambda` is indeed a constant since each of the letters in the expression represent constants.

This page will help you to connect the dots: <a href="https://www.intmath.com/differential-equations/4-linear-des-order-1.php">Linear DEs of Order 1</a>

To help you get started, and to make the writing easier, I would suggest you make these substitutions:

`z = ln(y(t))`

So the left hand side of (13) is now: `(dz)/(dt).`

`Q = lambda ln(y"*")`

Can you write what `P` is? And can you go on from there?

Re: How to transform the differential equation?

Dear Prof. Murray,thanks for your advice.
I will continue to do it later.
Would it be convenient for you to leave a contact number, email or Wechat ID?

X

Dear Prof. Murray,thanks for your advice.
I will continue to do it later.
Would it be convenient for you to leave a contact number, email or Wechat ID?

Re: How to transform the differential equation?

@zhangyhui: The idea of the forum is so that all readers can benefit from the exchange of questions, attempts and discussion. So feel free to keep asking questions here on the forum!

X

@zhangyhui: The idea of the forum is so that all readers can benefit from the exchange of questions, attempts and discussion. So feel free to keep asking questions here on the forum!

Re: How to transform the differential equation?

Thanks for your help.
This is what I have done.
`\frac{dz}{dt} = \lambda(z"*"- z)\ \ \ ` (1)
`\frac{dz}{dt} + \lambda z= \lambda \ln y"*" \ \ \ ` (2)

`\frac{dz}{dt} + P z= Q \ \ \ ` (3)
where `P = \lambda`,`Q =\lambda \ln y"*"" `.

`ze^{\int P dx} = \int(\lambda \ln y"*"e^{\int P dx}) dx +K`

`ze^{\lambda x} = \lambda \ln y"*" \frac{1}{\lambda}e^{\lambda x} +K`

`\ln y(t) = \ln y"*"+K e^{-\lambda x} `

But I can not still get Eq(14).
What else have I left out?
Thanks a lot!

X

Thanks for your help.
This is what I have done.
`\frac{dz}{dt} = \lambda(z"*"- z)\ \ \ ` (1)
`\frac{dz}{dt} + \lambda z= \lambda \ln y"*" \ \ \ ` (2)

`\frac{dz}{dt} + P z= Q \ \ \ ` (3)
where  `P = \lambda`,`Q =\lambda \ln y"*"" `.

`ze^{\int P dx} = \int(\lambda \ln y"*"e^{\int P dx}) dx +K`

`ze^{\lambda x} = \lambda \ln y"*" \frac{1}{\lambda}e^{\lambda x} +K`

`\ln y(t) = \ln y"*"+K e^{-\lambda x} `

But I can not still get Eq(14).
What else have I left out?
Thanks a lot!

Re: How to transform the differential equation?

Just for updates.
`\frac{dz}{dt} = \lambda(z"*"- z)\ \ \ ` (1)
`\frac{dz}{dt} + \lambda z= \lambda \ln y"*" \ \ \ ` (2)

`\frac{dz}{dt} + P z= Q \ \ \ ` (3)
where `P = \lambda`,`Q =\lambda \ln y"*"" `.

`ze^{\int P dt} = \int(\lambda \ln y"*"e^{\int P dt}) dt +K`

`ze^{\lambda t} = \lambda \ln y"*" \frac{1}{\lambda}e^{\lambda t} +K`

`\ln y(t) = \ln y"*"+K e^{-\lambda t} `

But I can not still get Eq(14).
What else have I left out?
Thanks a lot!

X

Just for updates.
`\frac{dz}{dt} = \lambda(z"*"- z)\ \ \ ` (1)
`\frac{dz}{dt} + \lambda z= \lambda \ln y"*" \ \ \ ` (2)

`\frac{dz}{dt} + P z= Q \ \ \ ` (3)
where  `P = \lambda`,`Q =\lambda \ln y"*"" `.

`ze^{\int P dt} = \int(\lambda \ln y"*"e^{\int P dt}) dt +K`

`ze^{\lambda t} = \lambda \ln y"*" \frac{1}{\lambda}e^{\lambda t} +K`

`\ln y(t) = \ln y"*"+K e^{-\lambda t} `

But I can not still get Eq(14).
What else have I left out?
Thanks a lot!

Re: How to transform the differential equation?

Seems pretty good what you've done.

You haven't used this part yet, though:

`y(0)` is income per effective worker at some initial date.

What's the value of `t` at the "initial date"?

X

Seems pretty good what you've done.

You haven't used this part yet, though:

`y(0)` is income per effective worker at some initial date.

What's the value of `t` at the "initial date"?

Re: How to transform the differential equation?

Thanks for your prompt reply.
Also deeply thanks for your inspiration.
`ln(y(0)) = \ln y"*" + K`
`K = ln(y(0)) - \ln y"*"`
substitute it, thus,
`\ln y(t) = \ln y"*"+(ln(y(0)) - \ln y"*") e^{-\lambda t} `
then (14) is derived.

Dear Prof.Murray, are you in Ox?
Sorry!

X

Thanks for your prompt reply.
Also deeply thanks for your inspiration.
`ln(y(0)) = \ln y"*" + K` 
`K = ln(y(0)) - \ln y"*"` 
substitute it, thus,
`\ln y(t) = \ln y"*"+(ln(y(0)) - \ln y"*") e^{-\lambda t} `
then (14) is derived.

Dear Prof.Murray, are you in Ox?
Sorry!

Re: How to transform the differential equation?

Great! Glad you sorted it out.

Actually, I live in Singapore. ^_^

X

Great! Glad you sorted it out.

Actually, I live in Singapore. ^_^

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