Explanation Theorem used: If f ( t ) is a periodic function of period T and that f ( t ) is piecewise continuous on [ 0 , ∞ ) , then, L { f ( t ) } exists for s > 0 and is given by the following definite integral. L { f ( t ) } = 1 1 − e − s T ∫ 0 T e − s t f ( t ) d t ⋯ ⋯ ( 1 ) Inverse Laplace Transform property: L − 1 ( e − a s G ( s ) ) = U ( t − a ) ⋅ g ( t − a ) Linear property of Laplace transforms: If L ( f ( t ) ) and L ( g ( t ) ) are exist, then L ( a f ( t ) + b g ( t ) ) = a L { f ( t ) } + b L { g ( t ) } . Linear property of inverse Laplace transforms: If L − 1 ( f ( t ) ) and L − 1 ( g ( t ) ) are exist, then L − 1 ( a f ( t ) + b g ( t ) ) = a L − 1 { f ( t ) } + b L − 1 { g ( t ) } . Laplace transform of the derivatives: L ( y ″ ) = s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) L ( y ′ ) = s Y ( s ) − y ( 0 ) Calculation: The given function is f ( t ) = { t − 1 , 0 ≤ t < 1 0 , 1 ≤ t < 2 and f ( t ) = f ( t − 2 ) if t ≥ 2 . From the definition of the function, f ( t ) is periodic with period T = 2 and is piecewise continuous on [ 0 , ∞ ) . Hence, L { f ( t ) } exists. Substitute T = 2 in ( 1 ) and proceed as below to obtain the Laplace transform of the given function. L { f ( t ) } = ( 1 1 − e − s ⋅ 2 ) ∫ 0 2 f ( t ) e − s t d t = ( 1 1 − e − 2 s ) [ ∫ 0 1 ( t − 1 ) ⋅ e − s t d t + ∫ 1 2 0 ⋅ e − s t d t ] = ( 1 1 − e − 2 s ) [ ∫ 0 1 t e − s t d t − ∫ 0 1 e − s t d t + 0 ] = ( 1 1 − e − 2 s ) ( [ t e − s t − s ] 0 1 − ∫ 0 1 e − s t − s d t − [ e − s t − s ] 0 1 ) Integrate further and evaluate the limits as follows. L { f ( t ) } = ( 1 1 − e − 2 s ) ( [ e − s − 0 − s ] − [ e − s t s 2 ] 0 1 − [ e − s − 1 − s ] ) = ( 1 1 − e − 2 s ) ( [ e − s − e − s + 1 − s ] − [ e − s − 1 s 2 ] ) = ( 1 1 − e − 2 s ) ( [ − 1 s ] − [ e − s − 1 s 2 ] ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ) Given differential equation is x ″ + 4 x = f ( t ) . Take Laplace transformation on both sides of the equation x ″ + 4 x = f ( t ) as follows. L { x ″ + 4 x } = L { f ( t ) } L { x ″ } + 4 L { x } = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ) Use Laplace transform derivative and substitute the initial value as follows. s 2 X ( s ) − s x ( 0 ) − x ′ ( 0 ) + 4 X ( s ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ) s 2 X ( s ) − 0 − 0 + 4 X ( s ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ) X ( s ) ( s 2 + 4 ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ) X ( s ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ( s 2 + 4 ) ) Rearrange and simplify further to obtain X ( s ) as shown below. X ( s ) = ( 1 1 − e − 2 s ) ( 1 − s − e − s s 2 ( s 2 + 4 ) ) = ( 1 1 − e − 2 s ) ( 1 s 2 ( s 2 + 4 ) − 1 s ( s 2 + 4 ) − e − s s 2 ( s 2 + 4 ) ) It is known that, the Taylor series expansion for 1 1 − x = 1 + x + x 2 + … = ∑ k = 0 ∞ x k . Substitute e − s T for x and obtain 1 1 − e − s T = 1 + e − s T + e − 2 s T + … = ∑ k = 0 ∞ e − k s T . Therefore, 1 1 − e − 2 s = 1 + e − 2 s + e − 4 s + … = ∑ k = 0 ∞ e − 2 k s . Substitute the above series expansion in X ( s ) . X ( s ) = ( 1 s 2 ( s 2 + 4 ) − 1 s ( s 2 + 4 ) − e − s s 2 ( s 2 + 4 ) ) ∑ k = 0 ∞ e − 2 k s = ∑ k = 0 ∞ e − 2 k s ( 1 s 2 ( s 2 + 4 ) − 1 s ( s 2 + 4 ) − e − s s 2 ( s 2 + 4 ) ) = ∑ k = 0 ∞ ( e − 2 k s s 2 ( s 2 + 4 ) − e − 2 k s s ( s 2 + 4 ) − e − ( 2 k + 1 ) s s 2 ( s 2 + 4 ) ) Next, evaluate x ( t ) as follows

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

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Chapter 8.4, Problem 60E

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The solution of initial value problem x″+4x=f(t), f(t)={t−1, 0≤t<10, 1≤t<2 and f(t)=f(t−2) if t≥2,x(0)=0.

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Chapter 8.4, Problem 59E

Chapter 8.4, Problem 61E

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Chapter 8 Solutions

Introductory Differential Equations

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The solution of initial value problem x ″ + 4 x = f ( t ) , f ( t ) = { t − 1 , 0 ≤ t &lt; 1 0 , 1 ≤ t &lt; 2 and f ( t ) = f ( t − 2 ) if t ≥ 2 , x ( 0 ) = 0 .

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The solution of initial value problem x″+4x=f(t), f(t)={t−1, 0≤t<10, 1≤t<2 and f(t)=f(t−2) if t≥2,x(0)=0.

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Chapter 8 Solutions

The solution of initial value problem x ″ + 4 x = f ( t ) , f ( t ) = { t − 1 , 0 ≤ t < 1 0 , 1 ≤ t < 2 and f ( t ) = f ( t − 2 ) if t ≥ 2 , x ( 0 ) = 0 .