Question

Consider the differential equation: -9ty" – 6t(t – 3)y' + 6(t – 3)y=0, t> 0. a. Given that yı(t) = 3t is a solution, apply the reduction of order method to find another solution y2 for which yı and y2 form a fundamental solution set. i. Starting with yi, solve for w in yıw' + (2y + p(t)yı)w = 0 so that w(1) = -3. w(t) = ii. Now solve for u where u = w so that u(1) = -5. u(t) = iii. Finally, write down y2 using the u that you found. yz(t) = b. Find the particular solution corresponding to the initial conditions y(9) = -1 and y'(9) = –2. Give your answer as y=.... Answer:

Answer 1

a Gireen D.E. is: -94²y" - 6+ (4-3) y +6 (-3)y 20, tro, & y(t) = 3t is a solution, we have to solve this by the reduction of onder method. let y = y, u), where 2 Y = BY, (t) = 3t , u=ult) DE Ô can be written as y" + 2/3 ( +-3) ayl - 23 ( + 2 y = 0 - ② which is of the form : " + Plt) g + Q(t) y 2o shtere PLU) = *(*). From @ we find y = 3t) u y = 34+ 3tu / Differentiating) y wir tot ) & y = 64't 3tu" substituting y", " & y values in ③ :- (6ult 3+") + 2 (4-3) (3u+ 3t ui) – 23 (+222) ostu zo - (3+)u" + (2x3 + 2 (+ 33 . (3+), u'=0 (4) which is of the form y, w't (24' + P (t)",) w zo. عدهمایی و .(ut = (هم اني

from (4), we get w'+ 33 w =0 - dw+ 2 + zo Integrating ; . ln(W) + 2 t = Alna W = Alst (5) where A is an integration (arbitrary) constant Gireen w(1) =-3. From (5) 2/2 6) 2t 6) W(+) = - 3€ (1-) = u(t) = - 3 0 3 (1-2) Integrating: - & ult) = -383 (edt +B => ult) = (2++B, B = ansitrany constant Ghreen U(1) =-5, in from (7), we get -5= 2 + B B = -1 u(t) = √ 1 99²311-72 197-

Also, from ②, cult) = Wit) = A e 2/3t. Integrating: - > uct) = - 3 Ae 3 +B , A, B are arbitrary constants. from ② , ie., Y = 7, u, we get "(t) = y(t) [B - 3 A e it ] /y(t) = B+(3+) - 2 Ateit Y(t) Differentiating was, to t we get y'(t) = 38 - Al estatēģt] (4) Given initial conditions: 4 (9) - 1 y(9)=-2 . i From (9), we get - 1 = 27B - 81 Aes & (11) - 2 = 3B + 45 AEG

ile (- ) A + 27B+1=0 (esse)A + 3B + 2 = 0 (cross multiplication) samo 481 => A=-B =-12 .. y(t) = (-)tet – 3 x 2 + 144) test - 2x27 y(t) = (17egt-23

in for the D.E. -9ty" -- 6t(+-3) y'+6(-3)=0,t%0, giren ý (t) = 34 is a solution, (i) Starting with. I, & solving for w in y, w't (24! + Plt) y ) w=0 so that w(1) =-3 we get wit) = -364) ci Then solring for u where u'aw so that u (2) = -5, we get uc)419e360-045]) dig vling us we get I as (to the particular solution connesponding to the initial conditions y()=-1 & y (8) = -2 is: y(t) = 1 3 [ 17 et * हमारा एग्जाम पैड प्राणियों का चबा राहत ह