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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 u003d 1; indeed, e u2212 2 x e u2212 x e x e 2 x u003d e u2212 2 x u2212 x u002b x u002b 2 x u003d e 0 u003d 1. In subtracting columns as indicated in the text, you get 1 0 0 0 u2212 2 1 3 4 4 u2212 3 u2212 3 0 u2212 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 u2212 3 0 0 7 2 16 u003d u002b 3 2 4 2 16 u003d 3 ue0a2 32 u2212 8 ue0a3 u003d 72. Problem Set 3 . 1 . Page 111 1 . Basis . A general solution is obtained by four successive integrations, y u2032u2032u2032 u003d c 1 , y u2032u2032 u003d c 1 x u002b c 2 , y u2032 u003d 1 2 c 1 x 2 u002b c 2 x u002b c 3 , y u003d 1 6 c 1 x 3 u002b 1 2 c 2 x 2 u002b c 3 x u002b c 4 , and y is a linear combination of the four functions of the given basis, as wanted. 5 . Basis . Setting y u2032u2032 u003d z , you have z u2032u2032 u002b 9 z u003d 0. A general solution is z u003d c 1 cos 3 x u002b c 2 sin 3 x . By integration you get y u2032 u003d u222b z dx u003d 1 3 c 1 sin 3 x u2212 1 3 c 2 cos 3 x u002b c 3 so that you obtain a general solution of the given ODE by another integration , y u003d u222b y u2032 dx u003d u2212 1 9 c 1 cos 3 x u2212 1 9 c 2 sin 3 x u002b c 3 x u002b 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 u2212 x are solutions of y u2032u2032 u2212 y u003d 0. Conclude that 1, e x , e u2212 x are solutions of the linear ODE y u2032u2032u2032 u2212 y u2032 u003d 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 u003d 1 e x e u2212 x 0 e x u2212 e u2212 x 0 e x e u2212 x u003d e x u2212 e u2212 x e x e u2212 x u003d 1 u2212 ue0a2 u2212 1 ue0a3 u003d 2. Conclude that the functions are linearly independent. Hence they form a basis of solutions of y u2032u2032u2032 u2212 y u2032 u003d 0. That e x e u2212 x u003d 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 u003d 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 y 1 u003d ln x , y 2 u003d ln x 2 , y 3 u003d ue0a2 ln x ue0a3 2 , y 1 u2212 1 2 y 2 u002b 0 y 3 u003d 0.
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