{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}