387_pdfsam_math 54 differential equation solutions odd

387_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 6.4 v2 (x) = = 1 x cos x(−2x3 ) dx = −16 8 x4 cos x dx 14 x sin x + 4x3 cos x − 12x2 sin x − 24x cos x + 24 sin x + c2 , 8 x cos x(−2x−1 ) 1 1 dx = v3 (x) = cos x dx = sin x + c3 , −16 8 8 where c1 , c2 , c3 are constants of integration, and we have used integration by parts to evaluate v1 (x) and v2 (x). Substituting these functions into (6.26) and simplifying yields yp (x) = − (x2 sin x + 2x cos x − 2 sin x) x + c1 x 4 x3 sin x (x4 sin x + 4x3 cos x − 12x2 sin x − 24x cos x + 24 sin x) x−1 + c2 x−1 + + c3 x3 + 8 8 = c1 x + c2 x−1 + c3 x3 − x sin x − 3 cos x + 3x−1 sin x . If we allow c1 , c2 , and c3 in the above formula to be arbitrary constants, we obtain a general solution to the original Cauchy-Euler equation. Thus, the answer is y (x) = c1 x + c2 x−1 + c3 x3 − x sin x − 3 cos x + 3x−1 sin x . 13. Since y1 Wk (x) = y1 . . . (n−2) y1 (n−1) y1 . . . y k −1 . . . y k −1 . . . ... ... (n−2) y k −1 (n−1) y k −1 0 yk+1 0 . . . 0 1 yk+1 . . . (n−2) yk+1 (n−1) yk+1 . . . yn . . . yn . . . ... ... (n−2) yn (n−1) yn , the k th column of this determinant consists of all zeros except the last entry, which is 1. Therefore, expanding Wk (x) by the cofactors in the k th column, we get Wk (x) = (0)C1,k + (0)C2,k + · · · + (0)Cn−1,n + (1)Cn,k y1 y1 = (1)(−1)n+k . . . y1 (n−2) . . . y k −1 . . . y k −1 . . . . . . y k −1 (n−2) yk+1 yk+1 . . . yk+1 (n−2) . . . yn . . . yn . . . . . . yn (n−2) 383 ...
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