Question

Using the classical method, solve

(D? + 2D y(t) = (D+1) f(t)

if the initial conditions are , , and if the input if f(t)=u(t)

Do not use convolution.

Answer 1

* Given that differential Equation (D?+20) y(t) = (0+1) FC) f (t) = UCt) 410)=2, 4(0)=1 i we have to find complete solution 1. To find complete solution (D?+2D) y() = (D+1) U(U) Dut) = 0 = DDE) + UCH) (D?+20) yu = UCH) = e. First we will find complimentry function D²+2D=0 DED+2) = 0 Roots are real f distinct so CF will be Cof= G+ C₂ e 2t For particular integral e pir= = ft) et Altl.eo p.C= 102420 e - (2D+2) 2D+2 P. f = Elts - Ult complete D=0, D= -2 D=0 soluton y(t) = CiftP.f. ulo=1 y(t) = G+ Ge2t + ULE) Initial Conditions y(o)=2= G + C₂ e 20 + ulo) 2= G+ G + 1 ath = 37 -

given another initial conditim Ý (O) = 1 2Cqe2t to It) = at to Ý (0) = 2 C eo 1 = 2 = 1/2 by putting value of Ca in Eqn (1) G+ G = 3/ G+ 1/2 = 3/2 = 3 2 - 2 = 32 (6 = 1) by putting ci & Ca in the complete solution we get NIN YCE = 1 + %++ voeten Note :- we can further simplity it but it depends on the what ? answer is required on in which form answer is require