Consider the following initial value problem. y′ + 5y = { 0 t ≤ 1 10 1 ≤ t < 6 0 6 ≤ t < ∞ y(0) = 4 (a) Find the Laplace transform of the right hand side of the above differential equation. (b) Let y(t) denote the solution to the above differential equation, and let Y((s) denote the Laplace transform of y(t). Find Y(s). (c) By taking the inverse Laplace transform of your answer to (b), the solution y(t) can be written in the form f (t) ?(t − 1) + g (t) ?(t − 6) + h(t). Enter the function f (t) into the answer box below. (d) Referring to part (c) above, enter the function g(t) into the answer box below. (e) Referring to part (c) above, enter the function h(t) into the answer box below. Consider the following initial value problem. y' + 5y { 0 t = 1 10 1st < 6 0 6 < t < oo y(0) = 4 (a) Find the Laplace transform of the right hand side of the above differential equation. (b) Let y(t) denote the solution to the above differential equation, and let Y((s) denote the Laplace transform of y(t). Find Y(s). (c) By taking the inverse Laplace transform of your answer to (b), the solution y(t) can be written in the form f(t) U(t – 1) + g (t) U(t – 6) + h(t). Enter the function f(t) into the answer box below. (d) Referring to part (c) above, enter the function g(t) into the answer box below. (e) Referring to part (c) above, enter the function h(t) into the answer box below.

Solh biven tsl y + 5y tello ostro het right hand side be alt) ts 710) = 4 0 0 istle bet (so 817) = 2[Bit)] = sest ait)dt riocistet tjoc.com 6 -st at t 6 vores dt e-st 1 -65 -Ceces] 11 -s 10 So taking the Laplace transformation both sides we get 2[%] + 5LIY] = lees] 3Y(S)- y (0) + 5 Y15) - loletes] Y(S) [s+5] - 4 pole sets) y(s) = 10 [es-eos sests) YES) lo es s(s+5) Scsts) + 4 s+5 So -65 + a sts c) 10 B s(s+5) s Sts 10 Alsts) + BS (A+B) S+SA SA = 10 → A=2 A+B= 0B = -2 1 11 S + L'( 559 ) [2 2 2 = 2 Elftal = eat Yes) = loets 2 SCS+5) S S+5 10 -5t 2- 2e S(545) And also we know that tas L -65 1o e SCS+5) s(3+5) -slt-1) FIS) I'[e-as Fis)] = H-a jutt-a) + 4 s+5 [[yis)] = {2-2e fuit) – (2-28 514-63 fulter) -st 4 e -slt-1) ylt) = 2/1-e { ult-1) - 2 1 - 2 514-6) (ults) -sti + 4e So we write ylt) = f(t) ult-1) + gt.) Ult-6) that) flt) = 281-esit g42 = -2hine -slt-6) C -5t hlt) = 40 .

Consider the following initial value problem. y&prime…

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Solh biven tsl y + 5y tello ostro het right hand side be alt) ts 710) = 4 0 0 istle bet (so 817) = 2[Bit)] = sest ait)dt rioc

1 11 S + L( 559 ) [2 2 2 = 2 Elftal = eat Yes) = loets 2 SCS+5) S S+5 10 -5t 2- 2e S(545) And also we know that tas L -65 1o

The method of Laplace Transforms is designed to work when the initial conditions are specified at t=0, but it can be adapted to deal with initial conditions at other values of t. This problem will show the method: y"(t) + 2y'(t) + y(t) = 4e + y(1)= '(1) = 0 a) Let y(t) = y(t+1); check that v(0) = y(1) and notice that v'(t)= y(t+1) and v"(t) =y"(t+1) b) Replace t by (t+1) on both sides of the differential equation, and then write the new equation in terms of the function v(t) c) What are the initial conditions v(O) and v'(0)? d) Solve the initial value problem for v(t) using Laplace transforms. e) Noticing that y(t) = v(t-1), what is the corresponding solution y(t)?

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Consider the following initial value problem. y′ + 5y = { 0 t ≤ 1 10...

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