hw5soln-corrected

# Cos x e 2x e 2x 1 u2 dx x c2 9 therefore

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Unformatted text preview: olution is ￿ ￿ ￿ ￿ 1 21 y = −t ln (t ) − + C 1 e −t + (t (ln (t ) − 1) + C 2 ) t e −t , 2 4 that is, 1 3 y = t 2 e −t ln (t ) − t 2 e −t + C 1 e −t + C 2 t e −t . 2 4 8 6 The equation to solve is ￿2 ￿ D − 2D + 2 y = e x sec (x ) . Solve the associated homogeneous equation. ￿2 ￿ D − 2D + 2 y ∗ = 0 (D − (1 − i )) (D − (1 + i )) y ∗ = 0 y ∗ = C 1 e x cos (x ) + C 2 e x sin (x ) Now, use variation of parameters. The general solution will be y = u 1 e x cos (x ) + u 2 e x sin (x ) , where ￿ u1 = = − y 1 f (x ) ￿ ￿ y1 y2 − y2 y1 − [e x sin (x )] [e x sec (x )] d d [e x cos (x )] d x [e x sin (x )] − [e x sin (x )] d x [e x cos (x )] =− =− e 2x tan (x ) e 2x cos (x ) (sin (x ) + cos (x )) − e 2x sin (x ) (cos (x ) − sin (x )) tan (x ) sin (x )2 + cos (x )2 = − tan (x ) ; u1 = ˆ − tan (x ) d x = − ln (cos (x )) + C 1 and ￿ u2 = = y 2 f (x ) ￿ ￿ y1 y2 − y2 y1 [e x cos (x )] [e x sec (x )] d d [e x cos (x )] d x [e x sin (x )] − [e x sin (x )] d x [e x cos (x )] e 2x e 2x = 1; = u2 = ˆ dx = x + C2. 9 Therefore the general solution is y = (ln (cos (x )) + C 1 ) e x cos (x ) + (x + C 2 ) e x sin (x ) , that is, y = e x (cos (x ) ln (cos (x )) + x sin (x ) + C 1 cos (x ) + C 2 sin (x )). 10 7 The equation to solve is equivalent to ￿ ￿ k 2 D+ y= m Make the substitutions w= to simplify to ￿ ￿ ￿ F k sin t. m m k F ; a= m m ￿ D 2 + w 2 y = a · sin (w t ) . (7.1) (7.2) Solve the associated homogeneous equation. ￿ ￿ D 2 + w 2 y∗ = 0 (D − i w ) ( D + i w ) y ∗ = 0 y ∗ = C 1 cos (w t ) + C 2 sin (w t ) (7.3) Now, make the judicious guess ￿ ￿ y = t α cos (w t ) + β sin (w t ) . Then so (7.4) ￿ ￿ ￿ ￿ D 2 y = 2 −w α sin (w t ) + w β cos (w t ) + t −w 2 α cos (w t ) − w 2...
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