Chapter 15

# Chapter 15 - 15 Integral Transform Method Exercises 15.1...

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15 Integral Transform Method Exercises 15.1 1. (a) The result follows by letting τ = u 2 or u = τ in erf( t ) = 2 π t 0 e u 2 du . (b) Using { t 1 / 2 } = π s 1 / 2 and the first translation theorem, it follows from the convolution theorem that erf( t ) = 1 π t 0 e τ τ = 1 π { 1 } t 1 / 2 e t = 1 π 1 s t 1 / 2 s s +1 = 1 π 1 s π s + 1 = 1 s s + 1 . 2. Since erfc( t ) = 1 erf( t ) we have erfc( t ) = { 1 } − erf( t ) = 1 s 1 s s + 1 = 1 s 1 1 s + 1 . 3. By the first translation theorem, e t erf( t ) = erf( t ) s s 1 = 1 s s + 1 s s 1 = 1 s ( s 1) . 4. By the first translation theorem and the result of Problem 2, e t erfc( t ) = erfc( t ) s s 1 = 1 s 1 s s + 1 s s 1 = 1 s 1 1 s ( s 1) = s 1 s ( s 1) = s 1 s ( s + 1)( s 1) = 1 s ( s + 1) . 5. From table entry 3 and the first translation theorem we have e Gt/C erf x 2 RC t = e Gt/C 1 erfc x 2 RC t = e Gt/C e Gt/C erfc x 2 RC t = 1 s + G/C e x RC s s s s + G/C = 1 s + G/C e x RC s + G/C s + G/C = C Cs + G 1 e x RCs + RG . 702

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Exercises 15.1 6. We first compute sinh a s s sinh s = e a s e a s s ( e s e s ) = e ( a 1) s e ( a +1) s s (1 e 2 s ) = e ( a 1) s s 1 + e 2 s + e 4 s + · · · e ( a +1) s s 1 + e 2 s + e 4 s + · · · = e (1 a ) s s + e (3 a ) s s + e (5 a ) s s + · · · e (1+ a ) s s + e (3+ a ) s s + e (5+ a ) s s + · · · = n =0 e (2 n +1 a ) s s e (2 n +1+ a ) s s . Then sinh a s s sinh s = n =0 e (2 n +1 a ) s s e (2 n +1+ a ) s s = n =0 erfc 2 n + 1 a 2 t erfc 2 n + 1 + a 2 t = n =0 1 erf 2 n + 1 a 2 t 1 erf 2 n + 1 + a 2 t = n =0 erf 2 n + 1 + a 2 t erf 2 n + 1 a 2 t . 7. Taking the Laplace transform of both sides of the equation we obtain { y ( t ) } = { 1 } − t 0 y ( τ ) t τ Y ( s ) = 1 s Y ( s ) π s s + π s Y ( s ) = 1 s Y ( s ) = 1 s ( s + π ) . Thus y ( t ) = 1 s ( s + π ) = e πt erfc( πt ) . By entry 5 in the table 8. Using entries 3 and 5 in the table, we have e ab e b 2 t erfc b t + a 2 t + erfc a 2 t = e ab e b 2 t erfc b t + a 2 t + a 2 t = e a s s ( s + b ) + e a s s 703
-10 -5 5 10 x -2 -1 1 2 y erfHxL erfcHxL Exercises 15.1 = e a s 1 s 1 s ( s + b ) = e a s 1 s s s ( s + b ) = e a s s + b s s ( s + b ) = be a s s ( s + b ) . 9. b a e u 2 du = 0 a e u 2 du + b 0 e u 2 du = b 0 e u 2 du a 0 e u 2 du = π 2 erf( b ) π 2 erf( a ) = π 2 [erf( b ) erf( a )] 10. Since f ( x ) = e x 2 is an even function, a a e u 2 du = 2 a 0 e u 2 du.

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• Fall '07
• Delahanty
• Sin, Integral transforms, cosh cosh

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