Question

(1 point) Use the Laplace transform to solve the following initial value problem: y" – 2y + 10y = 0 y(O) = 0, y' (O) = 3 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) = By completing the square in the denominator and inverting the transform, find yt) =

PLEASE MAKE SURE TO ANSWER ALL EMPTY BOXES SHOWN ON THE PROBLEM PLEASE AND THANK YOU

Answer 1

Sol y"-24't loyzo 416)=0, 420)=3 I lyy = YIS) formular Llyly = SY(S) - Y10) Lly) = 8² y(s)-S Y (0) - y'lo). Now from [sycs) -syro)=y\n] -2 [919-38] 41 014183] -) s? y's) – 3 - 2 syls) +10 Y CE) =0 Is Y15) – 2 5 718) +10 Y/-320. Y(s) 2 5²_2540) = 3. 27(s) S-25+10 3 (52-28+10) ylt) = Lily is) 3²-25410 ylt= • Llaff aliftus = (-{y's)) = 21 22 23 2lly(s)} = 27 (3. (5-1)² + 9 = 3. ER = 3. et t sinat et ainst 8-12 +32 Tylu=