394_pdfsam_math 54 differential equation solutions odd

394_pdfsam_math 54 differential equation solutions odd -...

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Chapter 7 7. For s> 2, L ± e 2 t cos 3 t ² ( s )= Z 0 e st e 2 t cos 3 tdt = Z 0 e (2 s ) t cos 3 = lim N →∞ ³ e (2 s ) t ((2 s )cos3 t +3sin3 t ) (2 s ) 2 +9 ´ ´ ´ N 0 µ = lim N →∞ e (2 s ) N [(2 s N N ] (2 s ) (2 s ) 2 = s 2 ( s 2) 2 . 9. As in Example 4 on page 353 in the text, we frst break the integral into separate parts. Thus, L{ f ( t ) } ( s Z 0 e st f ( t ) dt = 2 Z 0 e st · 0 dt + Z 2 te st dt = Z 2 te st dt . An antiderivative o± te st was, in ±act, obtained in Problem 1 using integration by parts. Thus, we have Z 2 te st dt = lim N →∞ ³¶ te st s e st s 2 · ´ ´ ´ N 2 µ = lim N →∞ ³ Ne sN s e sN s 2 + 2 e 2 s s + e 2 s s 2 µ = 2 e 2 s s + e 2 s s 2 = e 2 s 2 s + 1 s 2 · = e 2 s 2 s +1 s 2 · . 11. In this problem, f ( t ) is also a piecewise defned ±unction. So, we split the integral and obtain f ( t ) } ( s Z 0 e st f ( t ) dt = π Z 0 e st sin + Z π e st · 0 dt = π Z 0 e
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