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Unformatted text preview: C HEMICAL E NGINEERING 132A Professor Todd Squires Spring Quarter, 2007 Review for Midterm (A) Be able to look at a differential equation and tell things about it. 1) Ordinary differential equation or Partial Differential Equation? P t = 2 P x 2 + u P x (1) is partial because P depends on both x and t . However, P 2 dP dx + d 2 P dx 2- x 3 = 0 (2) is ordinary, because only derivatives with respect to x appear. 2) What order? d 2 P dx 2 + x = 0 (3) is second-order because the highest derivative is a second derivative, whereas y 5 dy dx + sin y = 0 (4) is first order. 3) Linear or nonlinear? x 4 dy dx + 2 sin( x ) y = 5 x 4 (5) is linear, because y and its derivatives never appear in a nonlinear function (e.g. ( y ) 2 , sin( y ), etc.). yy + sin( y ) = x (6) is nonlinear both because of the yy and because of the sin y . 4) Homogeneous or inhomogeneous? Easy way to tell if y = 0 solves the differential equation, it is homogeneous. Or are there any terms in the DE that do not contain y or its derivatives? y + 2 y = 4 x 3 (7) is inhomogeneous the 4 x 3 term does not depend on y , and so is an inhomogeneous term. y 3 y + sin( y ) sin( x ) y + 2 y = 0 (8) is homogeneous all terms depend on y , so that plugging in y = 0 does solve the equation. B) How to solve them? First off you do not know yet how to solve any PDEs. All we know are ODEs. 1) Second order ODEs you only know how to solve one kind, which are linear, second-order ODEs with constant coefficients. They can be homogeneous or inhomogeneous. Example: 2 y + 2 y + 5 y = 5 x (9) (Note this is inhomogeneous). First solve the homogeneous equation, by plugging in a trial solution y = e x (10) and solving for the s that work. In this case, this gives 2 2 + 2 + 5 = 0 , (11) which is solved by =- 4 4- 40 4 =- 4 - 36 4 =- 1 3 2 i (12) So the solution to the homogeneous equation is y = Ae x +3 ix/ 2 + Be x 3 ix/ 2 . (13) Note also that the i bits can also be expressed as sins and cosines: y = Ae x sin parenleftbigg 3 x 2 parenrightbigg + Be x cos parenleftbigg 3 x 2 parenrightbigg . (14) Weve done the homogeneous equation now how to find the solution to the inhomogeneous equation? This is called the particular solution. Easiest way undetermined coefficients. Basically try to guess the right answer. Here the inhomogeneous term is 2 x a first order polynomial. So try a particular solution that is also a first order polynomial: y p = Rx + S. (15) Plug this in y p = 0, y p = R . Plugging these in, we get 2 R + 5 Rx + 5 S = 5 x. (16) We must gather all the x terms and all the 1 terms (those with no x s). So the x terms are 5 R = 5 (17) or R = 1. The 1 terms give 2 R + 5 S = 0 , (18) or 5 S =- 2 R =- 2 , giving S =- 2 / 5....
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This note was uploaded on 07/12/2009 for the course CHE 132a taught by Professor Gordon,m during the Spring '08 term at UCSB.
- Spring '08
- Chemical Engineering