Solve the initial value problem below.using the method of Laplace transforms.y"-2y - 24y= 0, y (0)=1y(t) =Table of Laplace Transformsf(t)F(s)=L{f(t)}11S.S>01tt", n=1,2,...sin ktcos kty (0) =46(please an answer in terms of e.).,s>0s-an!sn+1ks²+kSs>0S>0S>0,s>0s²+k²¹sProperties ofLaplace Transforms£{1+g) = L(1)+(a)(c)= c() for any constant c£1)(8) B.Z(1)(B)-NO)£ (ſ^'^) (s) = ²£{f}(s) - st(0)-f(0)£ {p²^³} (x) = sº £{1}(s) - s^~¹4(0)-8²-² (0)²¹) (0)£* ¹{F1+F₂} = Z' ¹ (F1} •£¹ (F2)x"^ ¹ (CF) = CI" ¹ (F)

Solve the initial value problem below. using the method of Laplace transforms. y"-24"-24y = 0, y (0)=1 y(t)= Table of Laplace f(t) 1 t t", n=1,2,... sin kt cos kt y (0)=46 (please an answer in terms of e.). Transforms F(s)=L(f(t)} 1 1 n! s> S>0 S>0 S>0 k +k S 2+2 2+42 +5>0 S>0 Properties of Laplace Transforms (+g)-()+(a) (c)-c) for any constant c (1) (0) S.2(1)(B)-NO) 2) (s)- ²2(1)(s)-f(0)-f(0) (f) (a)(1)-^¯ ¹¶(0) - ^²√ (0) --- X" ¹ (CF) = CX" ¹ (F)

Solve the initial value problem below. using the method of Laplace transforms. y"-24"-24y = 0, y (0)=1 y(t)= Table of Laplace f(t) 1 t t", n=1,2,... sin kt cos kt y (0)=46 (please an answer in terms of e.). Transforms F(s)=L(f(t)} 1 1 n! s> S>0 S>0 S>0 k +k S 2+2 2+42 +5>0 S>0 Properties of Laplace Transforms (+g)-()+(a) (c)-c) for any constant c (1) (0) S.2(1)(B)-NO) 2) (s)- ²2(1)(s)-f(0)-f(0) (f) (a)(1)-^¯ ¹¶(0) - ^²√ (0) --- X" ¹ (CF) = CX" ¹ (F)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.3: Lines

Problem 30E

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