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Solve each of the problems of initial value, using the method of the Laplace transform

**Laplace transform** Solve each of the problems of initial value, using the method of the Laplace transform 1\. Y"-5y'+4y=e-2t ; y(0)=1, y'(0)=-1 2\. Y"+y= t sent ; and(0)=1, y'(0)=2

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y "- 5 y ' + 4 y = e^(2t); we apply the transform to both sides: L{y "} - 5 L {y '} + 4 L{y} = L {e^2t}; s^2 L{y} - s\*and(0) - y'(0) - 5 \[sL{y} - y(0)\] + 4 L{y} = 1 / (s-2); I no longer arising; replacement with the initial values: s^2 L{y} - s +1 - 5 \[sL{y} - 1\] + 4 L{y} = 1 / (s-2) s^2 L{y} - s +1 - 5sL{y} +5 + 4 L{y} = 1 / (s-2) (s^2 -5s +4) \*L{y} - s + 6 = 1 / (s-2); (s^2 -5s +4) \*L{y} = \[1 / (s-2)\] + s - 6; clearance L{y}: L{y} = {1 / \[(s-2)\*(s^2 -5s +4)\]} + \[s/(s^2 -5s +4)\] - \[6/(s^2 -5s +4)\]; clearance and: and=L^(-1){1 / \[(s-2)\*(s^2 -5s +4)\]} + L^(-1)\[s/(s^2 -5s +4)\] - L^(-1)\[6/(s^2 -5s +4)\]; As (s^2 -5s +4) is factorizable: 5+-√(25 - 16) / 2; (5+-3)/2; s=1; s=4: By space, I do the fractions partial of each reverse separately: 1°) 1 / \[(s-2)\*(s-1)\*(s-4)\] = A/(s-2) + B/(s-1) + C/(s-4); 1/\[(s-2)\*(s-1)\*(s-4)\] = \[2)\*(s-1)\*(s-4)\] 1 = A(s-1)(s-4) + B(s-2)(s-4) + C(s-2)(s-1); I give values to s: 1= TO\*1\*(-2) + 0 B + 0C; for s=2; A=(-1/2); 1 = 0A + B\*(-1)(-3) + 0C; for s=1; B=(1/3); 1 = 0A + 0B + C\*2\*3; for s=4; C=1/6; L^(-1) { (-1/2)\* \[1/(s-2)\] + (1/3)\*\[1/(s-1)\] + (1/6)\*\[1/(s-4)\]}; As the Transformed inverse of 1/(s-a) = e^(at): \## (-1/2)e^2t + (1/3)e^t + (1/6)e^4t 2°) s/\[(s-1)(s-4)\]; I will avoid the intermediate steps: A/(s-1) + B/(s-4); s= A(s-4) + B(s-1); 1 = A(-3)+0B; for s=1; A=(-1/3) 4= 0A +3B; for s=4; B=4/3; (-1/3)\[1/(s-1)\] + (4/3)\[1/(s-4)\]; Transformed in reverse order: \## (-1/3)e^t + (4/3)e^4t; 3°) \[6/(s^2 -5s +4)\] or: 6/\[(s-4)(s-1)\]; 6=A/(s-4) + B/(s-1); 6 = A(s-1) + B(s-4); 6=3A; for s=4; A=2; 6= -3B; for s=1; B=-2; 2\*\[1/(s-4)\] - 2\*\[1/(s-1)\]; Transformed in reverse order: \## 2e^4t - 2e^t; All together now, with y(t); y(0)=1; y '(0) =(-1): y(t) = (-1/2)e^2t + (1/3)e^t + (1/6)e^4t +(-1/3)e^t + (4/3)e^4t +2e^4t - 2e^t; \ #### y(t) =(21/6)e^4t - (1/2)e^2t-2e^t Then attempt the second.

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