{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

appendixL3Growth - Mathematical Appendix of Lecture 3 Part...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Mathematical Appendix of Lecture 3 Part IIB, Paper 2: Growth Dr. Tiago Cavalcanti Michaelmas 2010 1. Recall that the fundamental equation of the capital stock in intensive form is: ˙ ˜ k ( t ) = s ˜ k ( t ) α - ( δ + n + g ) ˜ k ( t ) , given ˜ k (0) > 0 . (1) This is a non-linear differential equation in ˜ k . The stationary equilibrium is ˜ k * = [ s ( δ + n + g ) ] 1 1 - α . Define x ( t ) = ˜ k ( t ) 1 - α . Therefore: ˙ x ( t ) = (1 - α ) ˜ k ( t ) - α ˙ ˜ k ( t ) ˙ ˜ k ( t ) = ˙ x ( t ) (1 - α ) ˜ k ( t ) - α . (2) Using (2) into (1), we have that: ˙ x ( t ) = (1 - α ) s - (1 - α )( δ + n + g ) x ( t ) . (3) This is a standard linear first-order differential equation in x ( t ). The solution of this equation is the sum of a homogenous solution, x h ( t ), and a particular solution, x p ( t ). (If you are not sure about this, check any mathematical book on first-order differential equation - Alpha Chiang, “ Fundamental Methods of Mathematical Economics ”, for instance, contains material on this.) The particular solution is the one in which ˙ x ( t ) = 0, or x ( t ) = x p : ˙ x ( t ) = 0 x p = s ( δ + n + g ) . The homogenous solution solves: ˙ x ( t ) + (1 - α )( δ + n + g ) x ( t ) = 0 . 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The solution is x h ( t ) = Ce - (1 - α )( δ + n + g ) t , where C is a constant that should be determined. The general solution of the differ- ential equation: x
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern