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examples-ch3

# examples-ch3 - MATH 152 Fall 2006-07 Applied Linear Algebra...

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MATH 152 Fall 2006-07 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology September 1, 2006

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Contents Table of Contents iii 1 Introduction 1 2 First-Order Differential Equations 5 3 Second-Order 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

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Chapter 3 Second-Order Linear Equations ¥ Example 3.1 (Fundamental solutions) Verify that the functions y 1 ( t ) = t 2 and y 2 ( t ) = 1 /t are solutions of t 2 y 00 - 2 y = 0 , t > 0 . (3.1) Solution By t 2 y 00 1 - 2 y 1 = t 2 (2) - 2( t 2 ) = 0 = y 1 = t 2 is a solution of Eq (3.1) , t 2 y 00 2 - 2 y 2 = t 2 ( 2 t 3 ) - 2( 1 t ) = 0 = y 2 = 1 t is a solution of Eq (3.1) . Furthermore, for any constants c 1 and c 2 , t 2 ( c 1 y 1 + c 2 y 2 ) 00 - 2( c 1 y 1 + c 2 y 2 ) = c 1 ( t 2 y 00 1 - 2 y 1 ) + c 2 ( t 2 y 00 2 - 2 y 2 ) = c 1 (0) + c 2 (0) = 0 . c 1 y 1 + c 2 y 2 is also a solution of Eq (3.1). Since y 1 /y 2 = t 3 6≡ constant, y 1 and y 2 are linearly independent and hence c 1 y 1 + c 2 y 2 is the general solution of Eq (3.1), the second-order linear differential equation. 2 ¥ Example 3.2 (Fundamental solutions) Verify that the functions y 1 ( t ) = 1 and y 2 ( t ) = t are solutions of yy 00 + ( y 0 ) 2 = 0 , t > 0 . (3.2) Solution By y 1 y 00 1 + ( y 0 1 ) 2 = (1)(0) + (0) 2 = 0 = y 1 = 1 is a solution of Eq (3.2) , y 2 y 00 2 + ( y 0 2 ) 2 = ( t 1 / 2 )( - 1 4 t - 3 / 2 ) + ( 1 2 t - 1 / 2 ) 2 = 0 = y 2 = t is a solution of Eq (3.2) . But, for any constants c 1 and c 2 , ( c 1 y 1 + c 2 y 2 )( c 1 y 1 + c 2 y 2 ) 00 + [( c 1 y 1 + c 2 y 2 ) 0 ] 2 = ( c 1 + c 2 y 2 )( c 2 y 00 2 ) + c 2 2 ( y 0 2 ) 2 = ( c 1 + c 2 t 1 / 2 )( - 1 4 c 2 t - 3 / 2 ) + c 2 2 ( 1 4 t - 1 ) = - 1 4 c 1 c 2 t - 3 / 2 6≡ 0 . c 1 y 1 + c 2 y 2 is, in general, not a solution of Eq (3.2). It is because Eq (3.2) is indeed a nonlinear equation. 2 39

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3. Second-Order Linear Equations ¥ Example 3.3 (Fundamental solutions) Let y 1 and y 2 be solutions of the homogeneous equation y 00 + p ( t ) y 0 + q ( t ) y = 0 (3.3) satisfying respective initial conditions y 1 (0) = 1 , y 0 1 (0) = 0; y 2 (0) = 0 , y 0 2 (0) = 1 . Find a solution satisfying the conditions y (0) = 3 , y 0 (0) = - 5 . Solution By the given initial conditions of y 1 (0) and y 2 (0), we know that y 1 and y 2 are linearly independent solutions of Eq (3.3). Then y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) is also a solution of the linear Eq (3.3). Because y 1 and y 2 are linearly independent, c 1 y 1 + c 2 y 2 is the general solution of Eq (3.3). Now 3 = y (0) = c 1 y 1 (0) + c 2 y 2 (0) = c 1 (1) + c 2 (0) , - 5 = y 0 (0) = c 1 y 0 1 (0) + c 2 y 0 2 (0) = c 1 (0) + c 2 (1) . Thus we have c 1 = 3 and c 2 = - 5. The solution is y = 3 y 1 - 5 y 2 . 2 ¥ Example 3.4 (Linear property) Given that y 1 = t + 3 4 is a solution of y 00 + 3 y 0 - 4 y = - 4 t, y 2 = te - 4 t is a solution of y 00 + 3 y 0 - 4 y = - 5 e - 4 t . Find a solution of y 00 + 3 y 0 - 4 y = t + 2 e - 4 t . Solution By linear property, the function - 1 4 ( t + 3 4 ) - 2 5 ` te - 4 t ´ = - 3 16 - t 4 - 2 t 5 e - 4 t is a solution of y 00 + 3 y 0 - 4 y = - 1 4 ( - 4 t ) - 2 5 ( - 5 e - 4 t ) = t + 2 e - 4 t .
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