Drorbn - User contributions [en]

12-267

2012-10-08T17:27:13Z

Syjytg:

12-267

2012-10-03T01:20:24Z

Syjytg:

Notes for 12-267-120910/0:25:00

2012-09-12T03:08:50Z

Syjytg:

It is great that I am able to know some physics on this course. [[User:Syjytg|Syjytg]] 23:08, 11 September 2012 (EDT)

Notes for 12-267-120910/0:00:57

2012-09-12T03:05:53Z

Syjytg:

I really like the way you introduce the so called "movie action scenes" in this course. [[User:Syjytg|Syjytg]] 23:05, 11 September 2012 (EDT)

12-267

2012-09-12T03:00:04Z

Syjytg:

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{{12-267/Navigation}}
==Advanced Ordinary Differential Equations==
===Department of Mathematics, University of Toronto, Fall 2012===

{{12-267/Crucial Information}}

===Text===
Boyce and DiPrima, [http://ca.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002451.html Elementary Differential Equations and Boundary Value Problems] (current edition is 9th and 10th will be coming out shortly. Hopefully any late enough edition will do).

===Further Resources===

* [http://www.math.toronto.edu/undergrad/ Undergraduate Information] at the [http://www.math.toronto.edu/ UofT Math Department]

* [http://www.artsandscience.utoronto.ca/ofr/calendar/crs_mat.htm Undergraduate Course Descriptions].

* Vitali Kapovitch's 2007 classes: [http://www.math.toronto.edu/vtk/267/ Spring], [http://www.math.toronto.edu/vtk/267fall07/ Fall].

* Also previously taught by T. Bloom, C. Pugh, D. Remenik.

* My {{Pensieve Link|Classes/12-267/|12-267 notebook}}.

{{Template:12-267:Dror/Students Divider}}

===Solving the complicated integral in the Brachistochroe integral===

integral sqrt((d-y)/y) dy
= integral sqrt(d y-y^2)/y dy
For the integrand sqrt(d y-y^2)/y, complete the square:
= integral sqrt(d^2/4-(y-d/2)^2)/y dy
For the integrand sqrt(d^2/4-(y-d/2)^2)/y, substitute u = y-d/2 and du = dy:
= integral (2 sqrt(d^2/4-u^2))/(d+2 u) du
= 2 integral sqrt(d^2/4-u^2)/(d+2 u) du
For the integrand sqrt(d^2/4-u^2)/(d+2 u), (assuming all variables are positive) substitute u = 1/2 d sin(s) and du = 1/2 d cos(s) ds. Then sqrt(d^2/4-u^2) = sqrt(d^2/4-1/4 d^2 sin^2(s)) = 1/2 d cos(s) and s = sin^(-1)((2 u)/d):
= d^2/2 integral (cos^2(s))/(d sin(s)+d) ds
For the integrand (cos^2(s))/(d sin(s)+d), substitute p = tan(s/2) and dp = 1/2 sec^2(s/2) ds. Then transform the integrand using the substitutions sin(s) = (2 p)/(p^2+1), cos(s) = (1-p^2)/(p^2+1) and ds = (2 dp)/(p^2+1):
= d^2/2 integral (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) dp
Simplify the integrand (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) to get (2 (p-1)^2)/(d p^4+2 d p^2+d):
= d^2/2 integral (2 (p-1)^2)/(d p^4+2 d p^2+d) dp
= d^2 integral (p-1)^2/(d p^4+2 d p^2+d) dp
= d^2 integral (p-1)^2/(d (p^2+1)^2) dp
= d integral (p-1)^2/(p^2+1)^2 dp
For the integrand (p-1)^2/(p^2+1)^2, use partial fractions:
= d integral (1/(p^2+1)-(2 p)/(p^2+1)^2) dp
= d integral 1/(p^2+1) dp-2 d integral p/(p^2+1)^2 dp
For the integrand p/(p^2+1)^2, substitute w = p^2+1 and dw = 2 p dp:
= d integral 1/(p^2+1) dp-d integral 1/w^2 dw
The integral of 1/(p^2+1) is tan^(-1)(p):
= d tan^(-1)(p)-d integral 1/w^2 dw
= d tan^(-1)(p)+d/w+constant
Substitute back for w = p^2+1:
= (d ((p^2+1) tan^(-1)(p)+1))/(p^2+1)+C
Substitute back for p = tan(s/2):
= 1/2 d (cos(s)+2 tan^(-1)(tan(s/2))+1)+C
Substitute back for s = sin^(-1)((2 u)/d):
= 1/2 (sqrt(d^2-4 u^2)+2 d tan^(-1)((2 u)/(d (sqrt(1-(4 u^2)/d^2)+1)))+d)+C
Substitute back for u = y-d/2:
= d (-tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d)))+sqrt(y (d-y))+d/2+C
Factor the answer a different way:
= 1/2 (-2 d tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d))+2 sqrt(y (d-y))+d)+C
Which is equivalent for restricted y and d values to:
= y sqrt(d/y-1)-1/2 d tan^(-1)((sqrt(d/y-1) (d-2 y))/(2 (d-y)))+C [[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)