This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 152 Fall 200607 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006 ii Contents Table of Contents iii 1 Introduction 1 2 FirstOrder Differential Equations 5 3 SecondOrder Linear Equations 39 4 Laplace Transform 83 5 Matrix 127 6 Systems of Linear Equations 135 7 Euclidean Vector 155 8 Eigenvalue and Eigenvector 177 9 Systems of Differential Equations 209 10 Orthogonality 231 iii Chapter 9 Systems of Differential Equations ¥ Example 9.1 (Converting higherorder equations to a system of firstorder equations) Rewrite the equation as a system of firstorder equations and find its corresponding initial value problem. ( u 00 + 0 . 125 u + u = 0 , u (0) = 1 , u (0) = 2 . Solution Let x 1 = u and x 2 = u , we have one firstorder equation x 1 = x 2 . By the assumption, we also have u 00 = x 2 . From the original equation, we have x 2 +0 . 125 x 2 + x 1 = 0, or x 2 = x 1 . 125 x 2 . Therefore the system becomes ( x 1 = x 2 , x 2 = x 1 . 125 x 2 . The initial conditions are ( x 1 (0) = u (0) = 1 , x 2 (0) = u (0) = 2 . 2 ¥ Example 9.2 (Converting a system of firstorder equations to a higherorder equation) Rewrite the system as a secondorder equation and find its corresponding initial value problem. 8 > < > : x 1 = x 2 2 x 1 , x 2 = x 1 2 x 2 , x 1 (0) = 2 , x 2 (0) = 1 . Solution From the first equation, we have x 2 = x 1 + 2 x 1 . So, x 2 = x 00 1 + 2 x 1 . Putting it into the second, we have x 00 1 + 2 x 1 = x 1 2( x 1 + 2 x 1 ). Therefore the secondorder equation is x 00 1 + 4 x 1 + 3 x 1 = 0 . Now consider the initial conditions. From the first equation, we have x 1 (0) = x 2 (0) 2 x 1 (0) = 1 2 · ( 2) = 5 . Therefore the initial conditions are x 1 (0) = 2 , x 1 (0) = 5 . 2 209 9. Systems of Differential Equations ¥ Example 9.3 (Modeling with differential system) Consider the two interconnected tanks shown in the figure. Tank 1 initially contains 30 gal of water and 25 oz of salt, while Tank 2 initially contains 20 gal of water and 15 oz of salt. Water containing 1 oz/gal of salt flows into Tank 1 at a rate of 1 . 5 gal/min. The mixture flows from Tank 1 to Tank 2 at a rate of 3 gal/min. Water containing 3 oz/gal of salt also flows into Tank 2 at a rate of 1 gal/min (from the outside). The mixture drains from Tank 2 at a rate of 4 gal/min, of which some flows back into Tank 1 at a rate of 1 . 5 gal/min, while the remainder leaves the system. Tank 1 Tank 2 Q 2 ( t ) oz salt 20 gal water Q 1 ( t ) oz salt 30 gal water 1 . 5 gal/min 1 oz/gal 1 gal/min 3 oz/gal 3 gal/min 1 . 5 gal/min 2 . 5 gal/min (a) Let Q 1 ( t ) and Q 2 ( t ) , respectively, be the amount of salt in each tank at time t . Write down differential equations and initial conditions that model the flow process. Observe that the system of differential equations is nonhomogeneous....
View
Full
Document
This note was uploaded on 09/30/2010 for the course MATH MATH152 taught by Professor Kcc during the Spring '10 term at HKUST.
 Spring '10
 KCC
 Math, Linear Algebra, Algebra, Equations

Click to edit the document details