Question

Chapter Section , ustion 17 Stics becobonded to look for

Answer 1

• Solution : As i can see in the picture 0 - we have given differential equation y (2d-15)y't (2² 152+50)y=0 - We need to finel '2' for which the solution of equation ③ tend to o as t oo, aro the value of " for which the solutions of 0 tend to as as t oo So, the auxiliary equation of ® is m² (22-15)gm+ (2²152750)=0 so m=-bI56 ²496 = (22-15) 1 J (22-15)²-4 (22 150+50) (2d-15) 11747225-600-924602-200

= (22-15) 15 2 so the roots of equation mi = 22 +10 = 2-5 ® - - S m₂ = 22-20=2-10 ® are so, the solutions of equation (2-5)+ Yi = a ex-lost ac are arbit Constanta 92=ge Case I: If a so auf Goo, then we a can take 2 is any real number for which is y=0 and Y2=0 Casa: I f G to and ę to, then yio as too ipl 2-550 1. 25 26 (-00,5) -

9.2 PPP similarly for Y o as to ipfo 2-1020 le 2210 =) de 100, 10) - € Therefore from equation ② auf 6 26(-0,5) Hem if take eats from (-0,5) then the solution tend Zero of 6. 2 S = Now, since y,= G e ya = Ca ek lot for imbranded folution of ③ we should I have to and lg to. then y, 900, as t-800 ilf 2-570 ie 275 and Y₂ ao as t o ilp 2710 Hence dE (10,oo) for unbounded solection of 6.