Question

(1 pt) Use the Laplace transform to solve the following initial value problem: y" +-6y' + 9y = 0 y0) = 2, y'(0) = 1 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation = 0 Now solve for Y(s) = and write the above answer in its partial fraction decomposition, Y(s) = sta + Y(s) = 2 Now by inverting the transform, find y(t) =

Answer 1

Sol Given yl_ 6ylt9y=0 Y(0)=2, 7(0/=) dfylly = 3²y(s)-sy(0)-yo)) Lefylz= SY(S)-Y(0) {y} - 6 Lofyty + 9 Lofy}=0/0 -> s’y (s) – Syo) - v'O) – 6 (SV(S) - Yo)) +9 L{x}=0 => sy 6) – 512) -|-6 (Sy(S) – 2) +996)=0 =) 3²Y(S) -6SY(s) +9 y(s)-257 +12=0 =>y052-65+9) = 25-13 => YOS? Asta >Y(s) = 2352 - © Taking partial fraction | YCS) = A +822) -) - 25-11 - A (5-3) + B (5-3) ² (5-3)? - 25-11 = AS – 3A +B - AS=25 -3A+B=-11. -) Y(s) = 25-11

- A=2, -3A+B=-1) -3(2)+B=-1) B=-1)+6 TA=2, B=5 substitute in ean ③ we get Y(S) = + (-5) Taking inverse laplace transformation 2 v0)3= 2414233-52166 L'{{a} = eat de og bestift Y(t) = 2034-5234648? = 20 3t - 58 3tt Ty(t) = 293t_st git

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