Question

+ Question 4. (4 marks) State the intervals in which there are sure to be solutions of the differential equation (x + 1)² y" – 3(x+1)y' + 3y = 0. Show that y = x + 1 is a solution to this differential equation, and then find a second lin- early independent solution using the Wronskian. Finally, determine the solution to the given equation that satisfies the conditions y(0) = 1 and y'(0) = 0.

Answer 1

Interval of existence, and showed y=x+1 is solution below. Using the ODE found the Wronskian explicitly. From there we found second solution y2 by solving its first order ode.

X+13%y" – 3(x+10y'+ 3y = 0 -7 y'' - (X+1) +12%. so, solution of this opE is sure to exist on R1{-1} sie intervals tco, u) U (1,0). * TS.T. y=x+1 is a solution y = 1 => y = o. put there in diff egn. 3y-3 y (8+) (x+12 3 3(x+1). - X + 1 84)2 so y = x+1 Stala'sfaj the ODE hence, is one of the solution with go) Let's call first solution to be y = x+1 & let seand solution is Ya. (linearly independ Wronskian, W(w)= yigh-gy differtiate لے are solh. 7 WCx)= y + y, y - yfy, +34 since, y,,92 3 gi -7 yi- 3 Yi (X4) Xri after putting they statisty Twins - Car So y dx WW)= 3 W(x). (x+1) y wkx) WCx)= e 4 (X+1) W(x) In WW)= Sx x cont.

be Xti e So, we calculated W(x) to dx est 3 ln (x+1). e. X + een(x+1} = (x +) )3 so, w(x) = (x+1)3 We want to find ya. (second solution) Wox ) = (X+1)² = ادا - ولا Bince y = x+. y = ! (x + 1)² = (x+1) 9₂ - Y₂ ya - 1 Y2 = (x+1) ². Xti We can solve this ODE to get ys 5 - 7 1 dx - ln(x+1) 1 e If= e X+1. #E using integrating factor multiply by ② by xH => --- & +12 423 X+1 Xt1 Y₂ = (x+1) - Xti २. # ys] => integrate XH

Now we have both solutions. using them and given intitial conditions following is the solution.

how, y = x+1 92= (x+1) २. We want to find solution statisfying y(o)= 1 y'(o)= 0. Jef ya C,CX+1) + C2 (x+103 2. y(0)=1 -) 15 ci + C2 2. y'= C1 + 2 530x+1)^ => o=c+ 3c2 2. y'0) = 0 .احتی همه sest Get C = 3% Ca = -1. 2 statisfying so, solution of ginen ODE y(0)=1 y/o) 20 is. y = - (x+; } 3(x+1) 2 2