Question

Solve the initial value problem. y" – 6y' +9y = 28(t – 1/3) cost y(0) = 0, y'(0) = 0

Answer 1

constants. Į so, Sol? y" - Gy' +94 = 28(+-) cost ; 7(0) = 0 = y(0) Taking Laplace transform, L18” - 48' + 3) = L 2 ( + - A) a +7 { Apply: Lfaflt) + bgcey} = al{f(t)}+bl{a(t)} where ad b are constants. Į so, L{x"} - 6L{'} +9L{x} = 21{$(+-) cost} {Apply 1{ y'} = 8L{x} - Y(0) ; 1{3"} = 82 L{x} – $36)=y() L{s(t-c) f(t)} = f(checs piso, $21283-80-0-6(8L{x} -0) +9L{x} = 2009(5) e. SPL{x} -681{x} +9L{x} = 2: {:.cos()= $} (82-68+9) L{y} = enge L{x} = enter 079 = 18-332 Taking Inverse Laplace transform, we get j(t) = ('{e *** encore}

8-21 1+1 !!= 1 2 = 1+1 Now, Let F(x) = -,2 ; 1"{F(x)} = f(t); 512=a} = {"{c} {E} {apply chain yang = the at fox ma41,2,3...) {1792} = t'est = text=fce Now; ["{e-7738. F(3)} calculation: ?"0 "se-cs Flu)? = ult-() f(+-) where ult-c) is Heaviside function & f(t) = ['{F(x)}}.so, L'{e-**. } = 417-4) (4-773) e368-74) L'{e7734 } = u(+-14)-(4-). elt-*) put value of LI { e-7/8. 1832} in equation to we put value get the solution of given initial value problem. t- y(t) = u(t-1/3):(t-13). e(3*-*) Ans.