Solutions for homework 8

Transcription

1 Solution for homework 8 Section. Baic propertie of the Laplace Tranform. Ue the linearity of the Laplace tranform (Propoition.7) and Table of Laplace tranform on page 04 to find the Laplace tranform of the function defined on the time domain. y(t) co t + 4 in 3t. For > 0, from the linearity of the Laplace tranform we have: L{ co t + 4 in 3t}() L{co t}() + 4L{in 3t}() ( + 9) + ( + ) ( + )( + 9). To tet the validity of Propoition ( + )(. + 9) (i) compute L(y )() for the given function, and (ii) compute L(y)() y(0) for the given function. Compare thi reult to that found in part (i) to verify that L(y )() L(y)() y(0). y(t) e 3t. (i) Uing the definition, the Laplace tranform of (e 3t ) i: L{y (t)}() L{(e 3t ) }() 3L{e 3t }() lim T T e (3+)t dt 3 lim T 0 e (3+)t (3 + ) T t e (3+)t dt e 3t e t dt (ii) Uing the Laplace tranform formula for the exponential we have: L(y)() y(0) L{e 3t }() e 3 0 ( 3)

2 9. Ue Propoition.,.4 and.7 to tranform the given initial value problem into an algebraic equation involving L(y). Solve the reulting equation for the Laplace tranform of y. y y e t. y(0). The Laplace tranform of the LHS i (ue the linearity of LT, Propoition., the initial condition): L(LHS)() : L(y y)() L(y )() L(y)() L(y)() y(0) L(y)() }{{} ( )L(y)() The Laplace tranform of the RHS i: L(RHS)() : L{e t }() With the notation Y () L(y)() we have then: ( )Y () + ( ) +. and therefore Y () ( + ) ( )( + ). 9. Ue Propoition. to find the Laplace tranform of the given function y(t) e t (t + 3t + 4). By Propoition., linearity of LT and formulae (.8), (.), we have L{e t (t + 3t + 4)}() L{t + 3t + 4} ( ( ) ) L{t }( + ) + 3L{t}( + ) + 4L{}( + )! ( + ) ( + ) ( + ) + 4( + ) ( + ) ( + ) 3.

3 Section.3 The Invere Laplace Tranform 3. Uing a table of Laplace tranform, much like uing a table of integral in a calculu cla, i not a traightforward a it would eem. Often, one ha to make adjutment to the given function in order to match a form in a table. It i the linearity of the Laplace tranform and it invere that make uch adjutment poible. For example, the form Y () 3 i not available in Table, but if we make the adjutment Y () 3, then by linearity, { } y(t) L 3 { } L 3 e 3 t. Ue thi technique to find the invere Laplace tranform of the function We can write and therefore we have Y () + 4. Y () y(t) in t. + L{in t}() 7. In Exercie 3 we ued the fact that L (αy ) αl (Y ). However, linearity in it more general form demand that L (αx + βy ) αl (X) + βl (Y ). The form Y () + +4 i not available in Table, but if we make the adjutment Y () , then, by linearity, { y(t) L } L co t + in t. + 4 ue thi technique to find the invere Laplace tranform of the function Y ()

4 Similarly we have Y () L{co t}() + L{in t}() L{3 co t}() + L{ in t}() L{3 co t + in t}(), and the invere Laplace tranform i: y(t) 3 co t + in t.. The terminology tranform pair i popular with engineer, and notation uch a y(t) Y () i ued to denote a tranform pair. For example e at a. Uing thi notation, if y(t) Y () i a tranform pair, the Propoition. tell u that e at y(t) Y ( a) i a tranform pair. For example, becaue Prpoition. tell u that co t + 4, e 3t co t 3 ( 3) + 4. Ue thi technique to find the invere Laplace tranform of the function Y () ( + ) 3. Uing the linearity of the invere Laplace tranform, Propoition., formula (.8) with n, we obtain: L { ( + ) 3 }(t) L { ( + ) 3 }(t) e t L { 3 }(t) e t L { 3 }(t) e t L { 3 }(t) e t t t e t.

5 9. Perform the appropriate partial fraction decompoition, and then ue the reult to find the invere Laplace tranform of Y () ( + )( ). Firt we find the partial fraction decompoition: ( + )( ) A + B A( + ) + B( ) + ( + )( ) and uing the ubtitution method, i.e., taking and, repectively we get: A 3, B 3, yielding ( + )( ) Then uing the linearity of the invere Laplace tranform, the formula with a and a, we obtain: L{e at }() a y(t) L { }(t) 3 L { }(t) 3 L { + }(t) 3 et 3 e t.

6 3 Section.4 Uing the Laplace Tranform to olve Differential Equation 7. Ue the Laplace tranform to olve the firt-order initial value problem y + 8y e t in t, y(0) 0. We compute the Laplace tranform of the LHS uing the linearity of the LT, formula (.) and the initial condition: L(y + 8y)() L(y + 8y)() L(y )() + 8L(y)() L(y)() y(0) +8L(y)() }{{} L(y )() ( + 8)L(y)() y(0) ( + 8)L(y)(). }{{} 0 The Laplace tranform of the RHS, by Propoition. or formula 6 in the Table (with a, b ), write: L{e t in t}() ( + ) +, and therefore the initial value problem reduce to the algebraic equation: i.e., ( + 8)L(y)() L(y)() Now we ue partial fraction ( + ) +, ( ( + ) + ) ( + 8). ( ( + ) + ) ( + 8) A B + C ( + ) + A[( + ) + ] + (B + C)( + 8) ( ( + ) + ) ( + 8) in order to find coefficient A, B, C uch that: With the ubtitution method: A[( + ) + ] + (B + C)( + 8). 8 A 37 and + i ( B( + i) + C ) (6 + i) ( B + C + ib)(6 + i) B + 6C B + i( B + C + 6B) 3B + 6C + i(4b + C)

7 equivalently So the Laplace tranfrom i { 3B + 6C 4B + 6C 0, B 37, C L(y)() 37( + 8) ( ( + ) + ) ( + ) ( + ) ( + ) ( + ) + and uing again Table for the invere Laplace tranform we obtain: y(t) 37 e8t 37 e t co t e t in t.. Ue the Laplace tranform to olve the econd-order initial value problem y 4y e t, y(0), y (0) 0. The Laplace tranform of LHS, uing (.) i: L(y 4y)() L(y ) 4L(y)() L(y)() y(0) y (0) 4L(y)() }{{}}{{} 0 ( 4)L(y)() +. The Laplace tranform of LHS, uing Table i: Therefore L{e t }() +. L(y)() ( + )( 4) 4 + ( + )( )( + ) A + + B + C +. Uing the ubtitution method: A( )( + ) + B( + )( + ) + C( + )( ) + yield A 3, B, C 4. Uing Table to get the invere Laplace tranform from L(y)()

8 yield y(t) 3 e t et 4 e t.. Ue the Laplace tranform to olve the econd-order initial value problem y y y e t, y(0), y (0) 0. The Laplace tranform of LHS, uing (.), (.) i: L(y y y)() L(y ) L(y )() L(y)() ( L(y)() y(0) y (0) }{{}}{{} 0 ( )L(y)() + ( + )( )L(y)() +. The Laplace tranform of LHS, uing Table i: Therefore L(y)() Uing the ubtitution method: L{e t }(). L(y)() y(0) }{{} ( + )( ) ( )( ) ( + )( ) ( + )( ) + 3 ( + )( ) A + + B + C ( ). A( ) + B( + ) + C( + ) + 3 ) L(y)() yield A 9, B 4 9, C 9. Uing Table to get the invere Laplace tranform from L(y)() ( ) ( ) 4( ) + 3 ( ) yield y(t) 9 e t 4 9 et + 3 tet.