solutions - MATH 2930: Quiz 1. Solutions October 29, 2009...

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Unformatted text preview: MATH 2930: Quiz 1. Solutions October 29, 2009 Problem 1 For each of the following statements, indicate whether its True or False and explain your reasoning. (a) The functions f ( x ) = e x , g ( x ) = cos x , h ( x ) = sin x are linearly independent on the real line. Solution. Definition 1. The functions f ( x ) , g ( x ) and h ( x ) are linearly independent on the real line, if the only constants c 1 , c 2 , c 3 , such that c 1 f ( x ) + c 2 g ( x ) + c 3 h ( x ) , are c 1 = c 2 = c 3 = 0 . Otherwise the functions are linearly dependent on the real line. Using just the definition is more or less convenient for proving that a given set of functions is linearly dependent. In that case one just needs to provide a set of constants c 1 , c 2 , c 3 , such that not all of them are zeros and c 1 f ( x ) + c 2 g ( x ) + c 3 h ( x ) . If we want to prove linear independence, then we would have to check for all possible values of the constants that the linear combination is not identically zero. The difficulty is that there are infinitely many possible values for the constants. Another method is more universal and requires calculation of the Wron- skian of our three functions: det e x cos x sin x e x- sin x cos x e x- cos x- sin x = e x sin 2 x + e x cos 2 x- e x sin x cos x + e x sin 2 x + e x sin x cos x + e x cos 2 x = 2 e x . 1 The Wronskian is non-zero, so the functions f ( x ) , g ( x ) , h ( x ) are linearly independent. (b) If the function y 1 ( x ) = x sin x is a solution of some linear homoge- neous equation with constant coefficients, then the function y 2 ( x ) = sin x is also a solution of the same equation. Solution. In this problem we have to make an analysis, when the function y 1 ( x ) = x sin x can be a solution of a linear homogeneous equation with con- stant coefficients. The rule says that each root of the characteristic equation gives us a solution, all of those solutions are linearly independent, and any linear combination of those solutions is also a solution of the given differential equation. The only way, how the function y 1 ( x ) = x sin x can be a solution, is when the characteristic equation has a pair of roots i repeated at least twice. Then the two linearly independent solutions, that correspond to the second pair of roots i , are x sin x and x cos x . The two solutions that correspond to the first pair of roots i , are sin x and cos x . As a result, we can conclude that if the function x sin x is a solution of some linear homogeneous equation with constant coefficients, then the functions x cos...
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solutions - MATH 2930: Quiz 1. Solutions October 29, 2009...

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