Question

Solve the following ode using Laplace transform: y' - 5y = f(t); y(0) - 1 t; Ost<1 f(t) = 0; t21

Answer 1

Boit?! Given y'-5y = f(t); y 10) = da) = It; osta + f(t) = t[uct) - u(t-1)] + b)ult -J = tulti - tult-1) = tults - (t-1+1) ultri) flt tult) -(t-1) ult-1) - ulta) y! - 54 = f(t) ay-59 tult) - (t-1) ult-I) a ult-) Laplace Transform Applying L{ y'} – 5L{y} = 2{tutt)} &L {{t-Just-i] - 2{ult-1)} => [946) – 410] —54 sz 12 e s → [Sy(4) - 1 3 – 540 = 1 12 es ( JE I (1-5) YO) = 1+ 1 / 2 82f1 Y(a)= 32/3-5) I-1 12 el(1-4) S²(6-5) T

Applying partial fractions confort 12+1 123-5) A + B 32 + C S 3-5 S²+1 3215-9) ASIS-5) + B (5-5)+C52 S²15-5) 52 (A+C 1.52 +5A +B)s -58 comparing comesponding Coleficients on both sides 52; 1= Atc =) J = ~ 11/ s &C =) C = 21 s': 0=-5A+B = -5A - 1 25 so: 5 50 A=-/ 25 븘 1-587 B = -1 5 .. A = -1 25 , B=/ 26 5 25 S²+1 545-5) B T/G + S 22 ta 5-5 - -1 S2+1 545-5) 255 + 26 552 2575-5)

By partial fractions -f A 3 + 543-5 + 3-5 -S 5?3-5) As Is-5) + B15-5) +C52 1 S215-5) & 1-5= (A+C) 52+(-5A +B)s - 5B Comp aring Corresponding Coefficients on both sides so: -5B = B = B = - 1 / 3 s': -= -5A+B= -5A -- -5A = -4/ =) -) A = + 4 25 52: O= Arc = A=-( =) C= *** B= 7/3, C = 11 25 :: A= -4 25 1-5 54-5) B C + S. S2 S-5

1-S 4 52(5-5) 255 4 2515-5 2 552 .: YO3) = [ Yes -425 f saya 26/25 t [45 / j2 4/25 Als Applying Inverse Laplace Transform YH =2"{ 903)} YH) = Lt -1/5 15 24/25 ? I + 82 4/25 /5 2 → YH) = (- 25 ist + 26 1 :) vex et5t) 518-met-> 25 4 25 7 syt-1) 4,51t- Is e

Ansurer is y(t) = 1 / 3 t/m t 5 t 26 25 est -( -- 25 Ilu 5 $ 2542-1)uto ult=1)