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Laplace Transform solved problems - Univerzita Karlova

Laplace Transform solved problems Pavel Pyrih May 24, 2012. ( public domain ). Acknowledgement. The following problems were solved using my own procedure in a program Maple V, release 5, using commands from Bent E. Petersen: Laplace Transform in Maple peterseb/mth256/docs/256winter2001 All possible errors are my faults. 1 Solving equations using the Laplace Transform Theorem.(Lerch) If two functions have the same integral Transform then they are equal almost everywhere. This is the right key to the following problems . Notation.(Dirac & Heaviside) The Dirac unit impuls function will be denoted by (t). The Heaviside step function will be denoted by u(t). 1. Problem. Using the Laplace Transform find the solution for the following equation . y(t) = 3 2 t t with initial conditions y(0) = 0. Dy(0) = 0.

1.1 Problem. Using the Laplace transform nd the solution for the following equation @ @t y(t) = 3 2t with initial conditions y(0) = 0 Dy(0) = 0 Hint.

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Transcription of Laplace Transform solved problems - Univerzita Karlova

1 Laplace Transform solved problems Pavel Pyrih May 24, 2012. ( public domain ). Acknowledgement. The following problems were solved using my own procedure in a program Maple V, release 5, using commands from Bent E. Petersen: Laplace Transform in Maple peterseb/mth256/docs/256winter2001 All possible errors are my faults. 1 Solving equations using the Laplace Transform Theorem.(Lerch) If two functions have the same integral Transform then they are equal almost everywhere. This is the right key to the following problems . Notation.(Dirac & Heaviside) The Dirac unit impuls function will be denoted by (t). The Heaviside step function will be denoted by u(t). 1. Problem. Using the Laplace Transform find the solution for the following equation . y(t) = 3 2 t t with initial conditions y(0) = 0. Dy(0) = 0.

2 Hint. no hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1 1. s Y(s) y(0) = 3 2 2. s s From this equation we solve Y (s). y(0) s2 + 3 s 2. s3. and invert it using the inverse Laplace Transform and the same tables again and obtain t2 + 3 t + y(0). With the initial conditions incorporated we obtain a solution in the form t2 + 3 t Without the Laplace Transform we can obtain this general solution y(t) = t2 + 3 t + C1. Info. polynomial Comment. elementary 2. Problem. Using the Laplace Transform find the solution for the following equation . y(t) = e( 3 t). t with initial conditions y(0) = 4. Dy(0) = 0. Hint. no hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t).

3 We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1. s Y(s) y(0) =. s+3. From this equation we solve Y (s). y(0) s + 3 y(0) + 1. s (s + 3). and invert it using the inverse Laplace Transform and the same tables again and obtain 1 1. + y(0) e( 3 t). 3 3. With the initial conditions incorporated we obtain a solution in the form 13 1 ( 3 t). e 3 3. Without the Laplace Transform we can obtain this general solution 1. y(t) = e( 3 t) + C1. 3. Info. exponential function Comment. elementary 3. Problem. Using the Laplace Transform find the solution for the following equation . ( y(t)) + y(t) = f(t). t with initial conditions y(0) = a Dy(0) = b Hint. convolution Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t).

4 We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain s Y(s) y(0) + Y(s) = Laplace (f(t), t, s). From this equation we solve Y (s). y(0) + Laplace (f(t), t, s). s+1. and invert it using the inverse Laplace Transform and the same tables again and obtain Z t y(0) e( t) + f( U1 ) e( t+ U1 ) d U1. 0. With the initial conditions incorporated we obtain a solution in the form Z t a e( t) + f( U1 ) e( t+ U1 ) d U1. 0. Without the Laplace Transform we can obtain this general solution Z. y(t) = e( t) f(t) et dt + e( t) C1. Info. exp convolution Comment. advanced 4. Problem. Using the Laplace Transform find the solution for the following equation . ( y(t)) + y(t) = et t with initial conditions y(0) = 1. Dy(0) = 0. Hint.

5 No hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1. s Y(s) y(0) + Y(s) =. s 1. From this equation we solve Y (s). y(0) s y(0) + 1. s2 1. and invert it using the inverse Laplace Transform and the same tables again and obtain 1 t 1. e + y(0) e( t) e( t). 2 2. With the initial conditions incorporated we obtain a solution in the form 1 t 1 ( t). e + e 2 2. Without the Laplace Transform we can obtain this general solution 1 t y(t) = e + e( t) C1. 2. Info. exponential function Comment. elementary 5. Problem. Using the Laplace Transform find the solution for the following equation . ( y(t)) 5 y(t) = 0. t with initial conditions y(0) = 2.

6 Dy(0) = b Hint. no hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain s Y(s) y(0) 5 Y(s) = 0. From this equation we solve Y (s). y(0). s 5. and invert it using the inverse Laplace Transform and the same tables again and obtain y(0) e(5 t). With the initial conditions incorporated we obtain a solution in the form 2 e(5 t). Without the Laplace Transform we can obtain this general solution y(t) = C1 e(5 t). Info. exponential function Comment. elementary 6. Problem. Using the Laplace Transform find the solution for the following equation . ( y(t)) 5 y(t) = e(5 t). t with initial conditions y(0) = 0. Dy(0) = b Hint. no hint Solution.

7 We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1. s Y(s) y(0) 5 Y(s) =. s 5. From this equation we solve Y (s). y(0) s 5 y(0) + 1. s2 10 s + 25. and invert it using the inverse Laplace Transform and the same tables again and obtain t e(5 t) + y(0) e(5 t). With the initial conditions incorporated we obtain a solution in the form t e(5 t). Without the Laplace Transform we can obtain this general solution y(t) = t e(5 t) + C1 e(5 t). Info. exponential function Comment. elementary 7. Problem. Using the Laplace Transform find the solution for the following equation . ( y(t)) 5 y(t) = e(5 t). t with initial conditions y(0) = 2. Dy(0) = b Hint.

8 No hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1. s Y(s) y(0) 5 Y(s) =. s 5. From this equation we solve Y (s). y(0) s 5 y(0) + 1. s2 10 s + 25. and invert it using the inverse Laplace Transform and the same tables again and obtain t e(5 t) + y(0) e(5 t). With the initial conditions incorporated we obtain a solution in the form t e(5 t) + 2 e(5 t). Without the Laplace Transform we can obtain this general solution y(t) = t e(5 t) + C1 e(5 t). Info. exponential function Comment. elementary 8. Problem. Using the Laplace Transform find the solution for the following equation 2. y(t) = f(t). t2. with initial conditions y(0) = a Dy(0) = b Hint.

9 Convolution Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain s (s Y(s) y(0)) D(y)(0) = Laplace (f(t), t, s). From this equation we solve Y (s). y(0) s + D(y)(0) + Laplace (f(t), t, s). s2. and invert it using the inverse Laplace Transform and the same tables again and obtain Z t y(0) + D(y)(0) t + f( U1 ) (t U1 ) d U1. 0. With the initial conditions incorporated we obtain a solution in the form Z t a + bt + f( U1 ) (t U1 ) d U1. 0. Without the Laplace Transform we can obtain this general solution Z Z. y(t) = f(t) dt + C1 dt + C2. Info. convolution Comment. advanced 9. Problem. Using the Laplace Transform find the solution for the following equation 2.

10 Y(t) = 1 t t2. with initial conditions y(0) = 0. Dy(0) = 0. Hint. no hint Solution. We denote Y (s) = L(y)(t) the Laplace Transform Y (s) of y(t). We perform the Laplace Transform for both sides of the given equation. For particular functions we use tables of the Laplace transforms and obtain 1 1. s (s Y(s) y(0)) D(y)(0) = 2. s s From this equation we solve Y (s). s3 y(0) + D(y)(0) s2 + s 1. s4. and invert it using the inverse Laplace Transform and the same tables again and obtain 1 1. t3 + t2 + D(y)(0) t + y(0). 6 2. With the initial conditions incorporated we obtain a solution in the form 1 1. t3 + t2. 6 2. Without the Laplace Transform we can obtain this general solution 1 2 1 3. y(t) = t t + C1 t + C2. 2 6. Info. polynomial Comment. elementary 10. Problem. Using the Laplace Transform find the solution for the following equation 2.