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.

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?

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?

@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!

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!

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!

`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"?

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!

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