Question

QUESTION 2 use to the following initial value problem (write fraction as (s- After Laplace Transform transform x" + 2x' +x=3, x(O)=0,x'(0)=1, you should get X(s)= S-2 2)/(S-4)(s+6) for (s-4)(8+6) -). Then, find x(t)= L-(x(s))= 5 -3t (write 5/6 by 6' e^{-3t} by e and sin(2t) or cos(3t) by sin(2t) or cos(3t)).

Answer 1

Given that x"(t)+2x'(t) +x(t) = 3, x(0) = 0, x'(0) = 1 Applying Laplace Transform on both sides L{x"(t)+2x'(t)+x(t) =L{3} L{x"(t))+22{x"(C)]+[{(t) = L{x"(c)} = sºL{z(t)} – sx(0) – x'(0) = s’L{x{t}}-1 L{x'(()) = sL{x()} - (0) =sL{x(t)} 3 8°L{x(t)) – 1+26L{x(t))+L{x(t)) = 3 1(10[+2+1] +1 +26+] [ +1] 1762***2+1) 918+1,2 1 X(s)=L(x(0)}= *(t)=1 B = + * 16+1) 3+5 s(s+1) 3+s s(s+1) (s +1) (s+1) 3+s Aſs+1)* + Bs(s+1)+Cs s(s+1) s(s+1) 3+s = A(s+1)* + Bs(s+1)+Cs Putting s=0= A=3 Putting s=-13C = -2 Comparing sa coefficients ,we get 0=A+B=B=-3 3+s 3 3 2 s(s+1)** 5 3+1 (5+1) 3+s = { (t)= 2 1 2(e)= 38"{}}-30*{1}-22 (5+1) 12-1 *(t)=3(1) -34*-2e-4 (2-1)! x(t)= 3-3e7-21