Question

Solve DE and Find path(time)

a) Y_t+2-2y_t+1+2y_t=1

y₀₌₁ y₁=4

b) Y_t+2-y_t+1+1/4y_t=1

Y₀=4 , Y₁=7

Answer 1

a) Solution ; y(t+2) - 2 y(t+1) + 2 y(«):1, . 7(0) = 1, y(i)=4. The equation can be written as, (DP-2D +2 ) y = 1. where , het Y= Aemt be the auxillary equation, then, m²-2m+2=0 e m= 2 + √4 - 4.2 2 = 21-a .. complementary function is, é* (quos tus sint). Iti Now, P. Io, 12-2D+2) 1 2 (4 =ţ (1+09-2014 = [- 0220 - 072037... ]2 - Ź So, general solection is y(t) = et (Geost +G sónt) + 1 Now, y(0)=1 37 1 =é (a coso tęsino) + 710-4 + el cost + C sín 1)+ £ 54 * .elt cost to sin 1) = 4 - 1/2 = 1 / 2 ¢ Plate - cosa) since Fatina ( - 20ss). Hence, the path, Y(u) a et (1 cost + á síni & usefnt ) +(Any)

b) y(++2) - y(t+1) +57(+)=1 , 460)=4, 4(1)-7. The equation can be written as, 1 (0²-D + 4) y =1., where af al =D Now, let y-emt be the auxillary equation of So, m?m+4=0. » (m-4):0. ► masa so, complementary function is, (a + Gt) e ² ***77?014. - 5-474 (1-2018 1:2 (1-20)725 = 2 (1 + 4D + higher onder of o). 1. = 2.1 so, the general equation is, y(t) = (a +ęte? + 2. Now, y 10)-4 4 = (4+0)ė +2 M, 4:2 and Y (1) at 7 7 = (2+G je²+2 one (+)een s as a = sele_ 2 Hence the path is from (2) : y(t) = [2 +150"-2)+] e+2 (Arus) (2)