#$&* course Mth 279 8/6 Query 31 Differential Equations*********************************************
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using, if necessary, the table in your text, find the Laplace transform of e^(2 t) cos(3 t). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(t) = e^(2 t) cos(3 t) gen form a^(at)cos(wt) = (s-a)/((s-a)^2 + w^2) a= 2 w=3 L{f(t)} = (s-2)/((s-2)^2 + 3^2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 10 / (s^2 + 25) + 4 / (s - 3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Kind of working backwards F(s) = 10 / (s^2 + 25) + 4 / (s - 3) find L^(-1){F(s)} take each part of function separately 10/(s^2+25) Laplace form of -> w/(s^2+w^2) Function ->f(t) =sin(wt) 2(5/(s^2+5^2) w=5 f(t) = 2sin(5t) 4/(s-3) Laplace form 1/(s-a) Function f(t)=e^(at) a=3 4(1/(s-3)) f(t) = 4e^(3t) f(t) = 2sin(5t)+4e^(3t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using, if necessary, the table in your text, find the inverse Laplace transform of e^(-2 s) / (s - 9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: L{f(t)} = e^(-2 s) / (s - 9) = e^(-2s)(1/(s-9)) e^(-2s) -> Laplace e^(-as)F(s) gen f(t) -> f(t-a)h(t-a) f(t) = f(t-2)h(t-2) 1/(s-9) -> gen Laplace 1/(s-a) a=0 gen f(t) = e^(at) f(t) = e^(9t) f(t)= e^(9(t-2))h(t-2) = L^(-1){f(t)} confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using, if necessary, the table in your text, find the inverse Laplace transform of 1 / (s + 1)^3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(s) = 1/(s+1)^3 gen Laplace - n!/(s-a)^(n+1) n=2 a=-1 gen f(t) => a^(at)t^n L^-1{2!/(s+1)^3} 2*L^-1{2!/(s+1)^3} = e^-1*t^2 L^-1{F(s)} = 1/2e^(-t)t^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 1
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Using, if necessary, the table in your text, find the inverse Laplace transform of (2 s - 3) / (s^2 - 3 s + 2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(s) = (2s-3)/((s-1)(s-2)) = A/(s-1) + B/(s-2) As-2A+Bs-B = 2s-3 A+B=2 -2A-B=-3 A=1 B=1 F(s) = 1/(s-1) + 1/(s+2) 1/(s-1) gen Laplace -> 1/(s-a) f(t) = e^(at) a_1=1 a_2 = 2 L^(-1){F(s)} = e^t + e^(2t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!