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Solve the given initial value problem using the method of Laplace transforms. 5, Ost≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7

Solve the given initial value problem using the method of Laplace transforms. 5, Ost≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve the given initial value problem using the method of Laplace transforms. 5, 0st≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7<t<14, 0, 14 <t Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. *** The solution has the form y(t) = p(t) + q(t)u(ta) + r(t)u(t-B), where u(t) is the unit step function. Let x <ß. Identify the values of a and B B = (Simplify your answers.)
Transcribed Image Text:Solve the given initial value problem using the method of Laplace transforms. 5, 0st≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7<t<14, 0, 14 <t Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. *** The solution has the form y(t) = p(t) + q(t)u(ta) + r(t)u(t-B), where u(t) is the unit step function. Let x <ß. Identify the values of a and B B = (Simplify your answers.)
f(t) 1 eat t, n=1,2,... sin bt cos bt eatin, n=1,2,... eat sin bt eat cos bt u(t-a) Ilab(t) £{f+g} = £{f} + £{g} F(s)= {f}(s) 1 £{eatf(t)} (s) = {f}(s-a) .s>0 S 1 s-a' n! 87+1.830 dn b s² + b² S s²+b² S>0 s>0 s>0 n! (s-a)n+1 b (s-a)² + b² s-a (s-a)2+b2 as S eas-e-bs S {cf} c{f} for any constant c = s>a s>a s>0 Properties of Laplace Transforms .s>a .s>0 £{f}(s) = sc{f}(s)-f(0) {f''(s) = s² {f}(s)-sf(0) -f'(0) {f(n)} (s) = s"{f}(s)-sn-1f(0)-s-2'(0)--(n-1)(0) c{tf(t)} (s) = (-1)- (c{f}(s)) ds £¯¹ {F₁ +F₂} = £¯ ¹ {F₁} + £¯¹ {F2} £¹(CF} = c£ ¹{F} e{f(t-a)u(t-a))(s) = eas F(s) ¹{e-as F(s)} (t) = f(t-a)u(t-a) {g(t)u(t-a)(s) = eas {g(t + a)(s)
Transcribed Image Text:f(t) 1 eat t, n=1,2,... sin bt cos bt eatin, n=1,2,... eat sin bt eat cos bt u(t-a) Ilab(t) £{f+g} = £{f} + £{g} F(s)= {f}(s) 1 £{eatf(t)} (s) = {f}(s-a) .s>0 S 1 s-a' n! 87+1.830 dn b s² + b² S s²+b² S>0 s>0 s>0 n! (s-a)n+1 b (s-a)² + b² s-a (s-a)2+b2 as S eas-e-bs S {cf} c{f} for any constant c = s>a s>a s>0 Properties of Laplace Transforms .s>a .s>0 £{f}(s) = sc{f}(s)-f(0) {f''(s) = s² {f}(s)-sf(0) -f'(0) {f(n)} (s) = s"{f}(s)-sn-1f(0)-s-2'(0)--(n-1)(0) c{tf(t)} (s) = (-1)- (c{f}(s)) ds £¯¹ {F₁ +F₂} = £¯ ¹ {F₁} + £¯¹ {F2} £¹(CF} = c£ ¹{F} e{f(t-a)u(t-a))(s) = eas F(s) ¹{e-as F(s)} (t) = f(t-a)u(t-a) {g(t)u(t-a)(s) = eas {g(t + a)(s)
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