wksht04sol

# wksht04sol - MATH 152 L1& L2 Applied Linear Algebra and...

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Unformatted text preview: MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 2004-05 ¥ Week 04 Worksheet Solutions : Second-Order Linear Equations Q1. (Linear property for second-order linear differential equations) (a) If y 1 ( t ) and y 2 ( t ) are two solutions of the linear homogeneous equation y 00 + cos ty = 0, and a 1 , a 2 are arbitrary constants then show the linear combination a 1 y 1 ( t ) + a 2 y 2 ( t ) is also a solution. Solution ( a 1 y 1 ( t ) + a 2 y 2 ( t )) 00 + cos t ( a 1 y 1 ( t ) + a 2 y 2 ( t )) = a 1 y 00 1 ( t ) + a 2 y 00 2 ( t ) + cos t · a 1 y 1 ( t ) + cos t · a 2 y 2 ( t ) = a 1 ( y 00 1 ( t ) + cos ty 1 ( t ) ) + a 2 ( y 00 2 ( t ) + cos ty 2 ( t ) ) = a 1 · 0 + a 2 · = . (b) If y 1 ( t ) is a solution of the linear nonhomogeneous equation y 00 +cos ty = 1 and y 2 ( t ) is a solution of the linear nonhomogeneous equation y 00 + cos ty = t , find a linear combination of y 1 and y 2 which is a solution of the linear nonhomogeneous equation y 00 + cos ty = 3- 2 t . Solution By linear property, the function 3 y 1 ( t )- 2 y 2 ( t ) is a solution of y 00 +cos ty = 3- 2 t . To show this, we suppose the function a 1 y 1 ( t )+ a 2 y 2 ( t ) to be the solution of y 00 + cos ty = 3- 2 t and attempt to determine the constants a 1 and a 2 . Now ( a 1 y 1 ( t ) + a 2 y 2 ( t )) 00 + cos t ( a 1 y 1 ( t ) + a 2 y 2 ( t )) = a 1 y 00 1 ( t ) + a 2 y 00 2 ( t ) + cos t · a 1 y 1 ( t ) + cos t · a 2 y 2 ( t ) = a 1 ( y 00 1 ( t ) + cos ty 1 ( t ) ) + a 2 ( y 00 2 ( t ) + cos ty 2 ( t ) ) = a 1 · 1 + a 2 · t = 3- 2 t. The last equality holds only if a 1 = 3 and a 2 =- 2. Q2. (Homogeneous equations with real constant coefficients) Consider the second-order linear homogeneous differential equation 6 y 00 + 5 y- 4 y = 0....
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wksht04sol - MATH 152 L1& L2 Applied Linear Algebra and...

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