Question

(2 points) Consider the following initial value problem, defined for t > 0: ' – 4y = f** (t – w) e4w dw, y(0) = -3. a. Find the Laplace transform of the solution. Y(s) = L {y(t)} b. Obtain the solution y(t). yt) =

Answer 1

ANSWER yl-4y = 4w -wle dw i 410)=-3 O a) Taking Laplace transform both sides we get Վա 2141) - 4219) awe dw w) D sys) - y0) - 446) = 2(t). d( e4t) (By convolution theorem, -- Syrs) + 3 -4 4S) X 52 S-4 - (5-4) y(s) + 3 s²(5-4) 15-4)*): 15-4 3 3 2 915) 2- s-4 - 4 - (5² - 4s)2 s-4

A + B soy S (5-4) S Als-4) +Bs 1 (A+B)S-YA - -4A=1; A+B=0 = B =- A A = -14 i B = 1/4 1 + 115-4) S(5-4) 2 +) 2 1652 16(5-4)2 165(5-4) (SIS-4)) ? + 8 4S 165² 415-4) 1615-4)2 + + 325 1615-4)2 165² 3215-4)

So 3 - + + Y(S) = 325 1652 1615-4)2 S-4 32(5-4) Taking Laplace inverse both sides weget - + - Yt) = + is d (152) 21 ) 16 52 32 2-) (54) 327 S-u 32 It t ! } tel ut to Чt Че - 3 e' + 32 32 so Y(A) teht + 16 + L 32 97 eut + 32