Question

Homework! This example illustrates how to proceed in the case of a second-order difference equation where the characteristic polynomial has two distinct real roots. The other two cases a repeated real root and a conjugate pair of complex roots - also work analogously to their DE counterparts. Try to solve these two linear recurrences. 1. 2. = 4(x,-1 -2,-2); *1 = 1, X2 = 3. (1, 3, 8, 20, 48,...) 2. In 2xn-1 - 4.xn-2; x1 = 1,22=3. (1,3,2,-8, -24,...)

Solve number 2 please

Answer 1

Note;

If the equation produces two distinct complex roots, m_{​​​​1} and m₂

in polar form

m_​​1=r ∠θ and

m₂ =r ∠(−θ)

Then

x_{​​​​​n} = r^{​​​​​​n} ( c_{​​​​​​1} cos(nθ) + c₂  sin(nθ) )

Is the solution

1-4х2 z-uxh Xn ; *,-1,2,=3 an 2xni + 4 an-2 O is Characteristic Equation m² am + 4 =0 2= ² 512 16 m=2 + 54 2 2 m 2+213 =ule It is 2 m;= 1+i53, -?u دکی ۶ -۱ converting them into polar form ohhom ro2 and O- TT 3 .. An=. 2" (4 cos ( mT) + C, sin (mun)) Now 2,-1 → 1- 2 CC, Cos (173) + y sim (1/3)) C,( ² ) +22( 3 ) V 2 C, + 53 e را - - Also E-?

( ( 27 ) ) 1 ( 2 ) د ) = 3 ( Cos ترام (1) + (5) 1) ت 3 + C - 2. ال + ) SX را « في ام Adding % and (ii) we obtian = ) 3 له S = دلا and hence ل + ,C 1 = (پتا) =) =رے 5 3 | : ہا xn= - 2 - (7) . ( n TT ولا Ans