Question

(1) Use the Laplace transform method to solve the initial value problem x + 2y. V=x+', (0) = 0 (0) -0. (Note that once you find either (t) or y(t), the other can be computed from the syste of ODE.) ISA - X-xo) = x +2Y ST-y(0) = X te 15.- 2 1 X(5-)-2Y=0 lo -2ts(s)

Answer 1

Sos differential equation consider the system of x = 2+28 y = xtet Take Laplace transform on both equation, we get L[u'] = b[n+a] L[8'] = L{xtel SL[1]-XCC) = 4[*] +2.13] Sh[8] - Yo) = 1[2] + Lief] SX-0 = X HRY SY -0 = x + - Oy= -S acol = 0 Leaf] = 4 -(3) - SX = X+27 Los sy=x+ - Q) from eancil, we get Sx-X = 2y => y = 5 (S-1)x substitute this in (?), we get s($cs-1x) = x +sta 7 (5+1)x - x = 5 >> (fu)-4)X = sa * (seps-')X = ca → (32-5-7)x=

=> x= (5-1) C53-5-2) = ţ (54)-(55) + (2) = 1 [jét - Of+cepty - Taking inverse Laplace transform on both sides, we get UC41 = þet-of Now substitute the value x in (3), we get y = $ (5-1)X = (5-1) X SUC533-2) 352-5-2 = -2.3ist 4-to-2 - (--- a) = - xnh (45) Taking muerse Laplace foans foom, we gee I'[ Y= eti? sinh ( 3#12) = ☆ COR_ē AJ = Y(+= { cept-et) Hence required solution to given fystem XCP) = še tref to op* J (H) = 1/3 ept 1/3 et Cans) is