16bf10lec15 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon () Math 16b Autumn 2010 1 / 25 Linear di erential equations A rst order linear di erential equation is a di erential equation of the form y + a ( t ) y = b ( t ) Thomas Scanlon () Math 16b Autumn 2010 2 / 25 Example Solve the di erential equation y + ty = Thomas Scanlon () Math 16b Autumn 2010 3 / 25 Solution Subtracting ty from both sides, we have the equation y =- t y In this case we can use the method of separation of variables. If y is constant, then ty ≡ y ≡ 0 so that y ≡ 0. Thomas Scanlon () Math 16b Autumn 2010 4 / 25 Solution, continued Otherwise, we may express the equation as y y =- t Let C = y ( ) . Integrating with respect to t , we have- 1 2 T 2 = Z T- tdt = Z T y dt y = ln | y ( T ) C | (As our solution must be continuous and cannot take the value zero, the signs of y ( T ) and C = y ( ) must agree. So, we may drop the absolute value bars.) Exponentiating both sides of this equation and multiplying by C , we obtain y ( T ) = Ce- 1 2 T 2 . Thomas Scanlon () Math 16b Autumn 2010 5 / 25 Another Example Solve the di erential equation y + y = 10 e- t Thomas Scanlon () Math 16b Autumn 2010 6 / 25 Solution In this case, we cannot apply the separation of variables technique. However, as e t is never equal to zero, the solutions to the original equation y + y = 10 e- t and to the equation e t y + e t y = 10 are the same. Using the chain rule, we see that d dt ( e t y ) = e t y + e t y Thus, our new di erential equation is d dt ( e t y ) = 10 Thomas Scanlon () Math 16b Autumn 2010 7 / 25 Solution, continued From the equation d dt ( e t y ) = 10 we integrate with respect to t . e T y ( T )- y ( ) = e t y ( t ) | t = T t = = Z T d dt ( e t y ) dt = Z T 10 dt = 10 T So, if we write C = y ( ) , then we have y ( T ) = 10 e- T T + Ce- T ....
View Full Document

This note was uploaded on 02/08/2011 for the course MATH 16B taught by Professor Sarason during the Fall '06 term at University of California, Berkeley.

Page1 / 24

16bf10lec15 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online