SSM-Ch03 - Chap. 3 Sec. 3.1 Higher Order Linear ODEs...

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Chap . 3 Higher Order Linear ODEs Sec . 3 . 1 Homogeneous Linear ODEs Example 5 . Basis , Wronskian . In pulling out the exponential functions, you see that their product equals e 0 u 1; indeed, e u 2 x e u x e x e 2 x u e u 2 x u x U x U 2 x u e 0 u 1. In subtracting columns as indicated in the text, you get 1 0 0 0 u 2 1 3 4 4 u 3 u 3 0 u 8 7 9 16 . You can now see the third-order determinant shown in the example. This determinant is simplified as indicated and then developed by the second row: 1 2 4 u 3 0 0 7 2 16 u U 3 2 4 2 16 u 3 u 32 u 8 U u 72. Problem Set 3 . 1 . Page 111 1 . Basis . A general solution is obtained by four successive integrations, y ±±± u c 1 , y ±± u c 1 x U c 2 , y ± u 1 2 c 1 x 2 U c 2 x U c 3 , y u 1 6 c 1 x 3 U 1 2 c 2 x 2 U c 3 x U c 4 , and y is a linear combination of the four functions of the given basis, as wanted. 5 . Basis . Setting y ±± u z , you have z ±± U 9 z u 0. A general solution is z u c 1 cos 3 x U c 2 sin 3 x . By integration you get y ± u U z dx u 1 3 c 1 sin 3 x u 1 3 c 2 cos 3 x U c 3 so that you obtain a general solution of the given ODE by another integration , y u U y ± dx u u 1 9 c 1 cos 3 x u 1 9 c 2 sin 3 x U c 3 x U c 4 . You see that it involves the given basis cos 3 x , sin 3 x , x , 1; it is a linear combination of these four functions with arbitrary coefficients. 7 . Linear independence . e x and e u x are solutions of y ±± u y u 0. Conclude that 1, e x , e u x are solutions of the linear ODE y ±±± u y ± u 0 with continuous (even constant) coefficients, so that you can apply Theorem 3. Calculate the Wronskian
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26 Ordinary Differential Equations (ODEs) Part A W u 1 e x e u x 0 e x u e u x 0 e x e u x u e x u e u x e x e u x u 1 u u u 1 U u 2. Conclude that the functions are linearly independent. Hence they form a basis of solutions of y UUU u y U u 0. That e x e u x u 1 is irrelevant since you have to consider linear combinations, not products of given functions. 9 . Linear dependence can often be shown by using a functional relation. In the present problem you have ln x 2 u 2 ln x and conclude that the functions are linearly dependent. Indeed, for instance, a linear combination with coefficients not all zero is, denoting the given functions by
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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SSM-Ch03 - Chap. 3 Sec. 3.1 Higher Order Linear ODEs...

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