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Lecture Notes in Differential Equations - Bruce E. Shapiro

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353<br />

Example 32.18. F<strong>in</strong>d the Laplace transform of the full-wave rectification<br />

of s<strong>in</strong> t,<br />

{<br />

s<strong>in</strong> t 0 ≤ t < π<br />

f(t) =<br />

(32.202)<br />

f(t − π) t ≥ π<br />

Us<strong>in</strong>g the formula for the transform of a periodic function,<br />

∫<br />

1 π<br />

F (s) =<br />

1 − e −πs s<strong>in</strong> te −st dt (32.203)<br />

0<br />

π<br />

1 1<br />

=<br />

1 − e −πs 1 + s 2 e−st (−s s<strong>in</strong> t − cos t)<br />

∣ (32.204)<br />

us<strong>in</strong>g (A.105) to f<strong>in</strong>d the <strong>in</strong>tegral. Hence<br />

1 1<br />

F (s) =<br />

1 − e −πs 1 + s 2 [e−πs (−s s<strong>in</strong> π − cos π)<br />

− e 0 (−s s<strong>in</strong> 0 − cos 0)] (32.205)<br />

1 1<br />

=<br />

1 − e −πs 1 + s 2 [e−πs (1) + (1)] (32.206)<br />

= 1 + e−πs<br />

1 − e −πs · 1<br />

1 + s 2 (32.207)<br />

Example 32.19. Solve y ′′ + 9y = cos 3t, y(0) = 2, y ′ (0) = 5.<br />

The Laplace Transform is<br />

0<br />

s 2 Y (s) − sy(0) − y ′ (0) + 9Y (s) =<br />

s<br />

s 2 + 9<br />

(32.208)<br />

Substitut<strong>in</strong>g the <strong>in</strong>itial conditions and group<strong>in</strong>g,<br />

(9 + s 2 )Y (s) − 2s − 5 = s<br />

s 2 + 9<br />

(32.209)<br />

rearrang<strong>in</strong>g terms and solv<strong>in</strong>g for Y (s),<br />

(9 + s 2 )Y (s) = 2s + 5 + s<br />

s 2 + 9<br />

(32.210)<br />

Y (s) = 2s + 5<br />

9 + s 2 + s<br />

(9 + s 2 ) 2 (32.211)<br />

= 2 ·<br />

s<br />

9 + s 2 + 5 · 1<br />

9 + s 2 + s<br />

(9 + s 2 ) 2 (32.212)

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