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examples152-ch3

# examples152-ch3 - MATH 152 Spring 2004-05 Applied Linear...

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Unformatted text preview: MATH 152 Spring 2004-05 Applied Linear Algebra & Differential Equations Worked Examples Dr. Tony Yee Department of Mathematics The Hong Kong University of Science and Technology February 15, 2005 ii Contents Table of Contents iii 1 Introduction 1 2 First-Order Differential Equations 5 3 Second-Order Linear Equations 39 iii 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 > . (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) = . ∴ 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 ) 2 = 0 , t > . (3.2) Solution By y 1 y 00 1 + ( y 1 ) 2 = (1)(0) + (0) 2 = 0 = ⇒ y 1 = 1 is a solution of Eq (3.2) , y 2 y 00 2 + ( y 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 ) ] 2 = ( c 1 + c 2 y 2 )( c 2 y 00 2 ) + c 2 2 ( y 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≡ . ∴ 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 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 + q ( t ) y = 0 (3.3) satisfying respective initial conditions y 1 (0) = 1 , y 1 (0) = 0; y 2 (0) = 0 , y 2 (0) = 1 . Find a solution satisfying the conditions y (0) = 3 , y (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) = c 1 y 1 (0) + c 2 y 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 ....
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examples152-ch3 - MATH 152 Spring 2004-05 Applied Linear...

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