Explanation Given: The integral equation is f ( t ) = e t + e t ∫ 0 t e − τ f ( τ ) d τ . Concept used: “The formula for the Laplace of e k t is defined below. L { e k t } = 1 s − k The formula for the inverse Laplace of 1 s − k is defined below. L − 1 { 1 s − k } = e k t The convolution theorem of inverse Laplace is mention below. L − 1 { f ∗ g } = L − 1 { f ( t ) } L − 1 { g ( t ) } = F ( s ) G ( s ) ” Calculation: Consider the function is f ( t ) = e t + e t ∫ 0 t e − τ f ( τ ) d τ . Now, move the term into the integral as given below. f ( t ) = e t + ∫ 0 t e t − τ f ( τ ) d τ Take the Laplace transformation on both side of above equation. L { f ( t ) } = L { e t } + L { ∫ 0 t e t − τ f ( τ ) d τ } The Laplace transformation is simplified as follows: L { f ( t ) }

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 4.4, Problem 54E

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The solution of the integral equation f(t)=et+et∫0te−τf(τ)dτ.

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Chapter 4.4, Problem 53E

Chapter 4.4, Problem 55E

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1. Matrix Operations Given: A = [ 33 ]A-[3-321] -3 B = [342]B-[3-41-2] (a) A² A2 Multiply A× A: -3 = (3 x 32x-3) (3 x 22 x 1) | = |[19–63 |-9-3 -6+21] = A² = 33 33 1-3×3+1x-3) (-3×2+1x1) [12]A2=[3-321][3-321]=[(3×3+2x-3)(-3×3+1x-3)(3×2+2×1)(-3×2+1×1)]=[9-6-9-36+2-6+1 ]=[3-128-5] (b) | A ||A| Determinant of A | A | (3 × 1) (2 x-3)=3+ 6 = 9|A|=(3×1)-(2x-3)=3+6=9 (c) Adjoint of A Swap diagonal elements and change sign of off-diagonals: A = [33], so adj (A) = |¯²]A=[3-321], so adj(A)=[13–23] -3 (d) B-¹B-1 First find | B ||B|: |B | (3x-2)- (1 × -4) = -6 + 4 = −2|B|=(3x-2)-(1x-4)=-6+4=-2 Then the adjoint of B: adj (B) = [² 3 adj(B)=[-24-13] Now, B-1 1 = |B| · adj (B) = 1 [²¯¯³¹³] = [2₂ B 0.5 |B-1=|B|1-adj(B)=-21[-24-13]=[1-20.5-1.5] 2. (a) Matrix Method: Solve (2x + 3y = 6 (2x-3y=14 {2x+3y=62x-3y=14 Matrix form: 22 33-22 = [223-3][xy]=[614] Find inverse of coefficient matrix: Determinant: | M | (2x-3) - (3 x 2) = -6 -6 = -12|M|=(2x-3)-(3×2)=-6-6=-12 Adjoint: adj(M) = [3]adj(M)-[-3-2-32] So…

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The solution of the integral equation f ( t ) = e t + e t ∫ 0 t e − τ f ( τ ) d τ .

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The solution of the integral equation f(t)=et+et∫0te−τf(τ)dτ.

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Chapter 4 Solutions