Question

#3 (50 pts) Use the Laplace transform (including all tables and theorems) to solve the initial value problem. y"- 4 y'= 6e" – 3e", y(0) = 1, y '(0)=-1

Answer 1

y" _44'=6 est set Y(0) 21, y'(o) = -1 Apply Laplace transform both sides, 2{y" _ay'} =2&6et_3 et} AL{y } - 4L{y'} = {belt} -2{3 ét} 6 3 -Y60) - 2400) - Y'(o) - 4[*Y») - 460] +1 >>Y10) - Ati-4240) + 4 int15 (1-3)(8 +1) 56-40Y(~) = s-5 t 3 st15 (5-3) (0+1) » (0-9) Y(2) - 8"-282-38 -5 $+108 +15 +30 +15 (0-3)(n+1) 0-70? +100+30 ols-4) Y(0) = (-3) (0+1) Y(0) = out volt il- (170)(8-4) (6-0) .: 8(t) = ["{2-20 ☆ _70² ton +30 (8-4) (0-3) (sti) -} Let ²-70²tion +30 А B + t o +1 no-9)(2-3) (+1) 3) 6-78° +102 +30 = A(n-4) (0-3) (+1) + BA (0-3710 +1) + C 810-4) (0+1). +Ds (2-4) (2-3) Pet, sco. 5 ::12A = 10 > A= pusti 20B = 2 2 B = Llo

Put, oe. .. -12 C = 24 PCs - 2 Put, p=-1 - 2 ) = ( 2 ) 2 = - 5 ) (는) t t t 0-4 ot! 2 -7 + 10 2 + 3 0 po - 4)(o - )(o +1) -4 2 LY - + . (e) = L 10(-0 20 5(+) 7 2L SI ?.. 유 + {} 부하 1 4 041 4t 5 을 -2et et