Question

Problem 5. (20 pts) Let ER be a positive real number and consider the damped system modeled by the following second-order differential equation: y"(t) + yy' (t) + 25y(t) = 0, (a) Show that the long-term behaviour of all solutions is independent of y. (b) For which values of ye R+ does the above differential equation have oscillating solutions ? (i.e. solutions with infinitely many zeroes.) (c) Classify the above damped system into underdamped, critically damped and overdamped in terms of y, and qualitatively draw solutions corresponding to each of these three cases. (c) Find all solutions to the above differential equation for y = 10. (d) For y = 10, let yı(t) be a solution such that yı(0) = 0 and y11) = 1, and yz(t) a solution such that y(0) = 1 and y,0) = 0. Compute the ration yı(t)/y2(t) for t oo and deduce which solution converges faster to zero.

Answer 1

Answer Let YER. be a real number. positive y" (+) + r y' (t) + 2546)=0. using characteristic Equation ش مع užtru +25=0. u= -11/12-100 u= rts-loc 100)= (-r-12-100 t + Be 2 y(t) = AS yet) - ette Faerite celor Bevo-100 C4) For ttoo Solutions are yet) to essentially independent of r. be Y EIRT IF r < 1oo. r< ricco = r < lo. As the power was imaginary.

Oscilatory solutions we get The above damped System Underdamped, critically damped Over damped. In terme of o. into and The characteristic Equation u=-TI V pr - 100 2 r=b/m and 25= k/m. so, u= ab Ibr-unk 2m. where mis mass of The System Case- 1: If b' < 4mk. under damping that Here conditan will where r < 100. case 2: of br > unk

damping then Here condition will over where th 2100. Case 3: Af brume. Here condition will be where m= 1000 critical damping. three cass, a gaph Corresponding to the shown below. is over damping critical damping . L under damping .

a to to the t To find all solutions differential Equation =b 1:10=-5. solution The ). is y(t): (A + Bté that V = 10 let y, (t) be a solution 4,(0)=0 Y, C1 ) = 1. YLt). y.co)- ' & Ye (0)=0. By competing y, Ct1/42Ct). to t-too. for Y CO) =O , y(i)=). 4,Co) = A =0 9,(1) = Besar =B=es. .: 4, (t) = (est jest

1.4,(t) = t.es Cint). - 4,Co)-1 Y,'(0) -0. 4,(0) - Aol y,'co) = (A+B+ )(-5)est + eStB = 0 s + B = O. 2. B= 5 -@) .: y, (t) = (1+st)est From we get Ego q qq ③ 4, (t) = estétt. Y. (f) (itstjext =est 1+5t es is for t o the to zero approximately Equal to 148. YeCt) solution will convere fastly, - y .