nagle_differential_equations_ISM_Part51

# nagle_differential_equations_ISM_Part51 - Chapter 7 Next we...

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Chapter 7 Next, we consider the initial value problem I 00 ( t ) + 8 I 0 ( t ) + 41 I ( t ) = 0 , I (0) = 2 , I 0 (0) = - 8 for the corresponding homogeneous equation. Its characteristic equation, r 2 +8 r +41 = 0, has roots r = - 4 ± 5 i , which give a general solution I h ( t ) = e - 4 t ( C 1 cos 5 t + C 2 sin 5 t ) . Next, we ﬁnd constants C 1 and C 2 so that the solution satisﬁes the initial conditions. Thus, we have 2 = I (0) = C 1 , - 8 = I 0 (0) = - 4 C 1 + 5 C 2 C 1 = 2 , C 2 = 0 , and so I k ( t ) = 2 e - 4 t cos 5 t . Finally, I ( t ) = [ h * ( e/ 10)] ( t ) + I k ( t ) = 1 50 t Z 0 e - 4( t - v ) sin [5( t - v )] e ( v ) dv + 2 e - 4 t cos 5 t. 32. By the convolution theorem, we get L ± 1 * t * t 2 ² ( s ) = L{ 1 } ( s ) L{ t } ( s ) L ± t 2 ² ( s ) = 1 s · 1 s 2 · 2 s 3 = 2 s 6 . Therefore, 1 * t * t 2 = L - 1 ³ 2 s 6 ´ ( t ) = 1 60 L - 1 ³ 5! s 6 ´ ( t ) = t 5 60 . 34. Using the commutative property (4) of the convolution and Fubini’s theorem yields ( f * g ) * h = ( g * f ) * h = t Z 0 ( g * f )( t - v ) h ( v ) dv = t Z 0 t - v Z 0 g ( t - v - u ) f ( u ) du h ( v ) dv = t Z 0 t - v Z 0 g ( t - v - u ) f ( u ) h ( v ) dudv = t Z 0 t - u Z 0 g ( t - u - v ) h ( v ) dv f ( u ) du = t Z 0 ( g * h )( t - u ) f ( u ) du = ( g * h ) * f = f * ( g * h ) . 246

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Exercises 7.8 36. Let G ( t ) := t Z 0 v Z 0 f ( z ) dz dv . Clearly, G (0) = 0. By the fundamental theorem of calculus, G 0 ( t ) = t Z 0 f ( z ) dz , G 0 (0) = 0 , G 00 ( t ) = f ( t ) . Therefore, by Theorem 5 in Section 7.3, we get
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## This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part51 - Chapter 7 Next we...

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