Question

a)
y" - y - 12y = 0
; and

b) ; and

c) ; and $y'_{(0)}=-3$

Answer 1

@ you - 4* –12y =o , 7(0) = 3 and y'(o)=5 cso Let y = emx be a solution of y = mene y" = ma enne So from 0 (ma _m-12) ent=0 1 m² _m-12=0 since ent never vanishes -4m +3m-12 = 0 = m(m-4) + 3(m-4)=0 = (m-4) (m + 3) =0 i M = -3,4 So the solution of ② is y = A € 32 + Beute @ y = -3A 232 +48 € -> 760)=3 in ☺, ie, 1-0, y = 3 3 = A +8 -> y' *)-5. From 3, 5 = -3A +43 > @x3 +6 = 9+5 =(3A+38) + (-3A +43) S 14 = 7B B=2 from 6, 3= Ata = A=1. Thus the solution of ② is y = ē3% 24x

Ⓡ y“ – 64'+8y=0 , 7(0) = 1, Y CO)=6 In this case, we have m-6m +8=0 m2 - 4m -2m +8=0 = m (m-4) – a(m-4)=0 (m-2) (m-4)=0 1. Thus, m=2,4 : The solution of 6 is - y = A ear toela y' = 2 A 62% + 4 Bett @ Put x=0, Y=1 in ②, I= A+B @ Put x = 0, y =6 in 6 = 24+48 + 3 = A +2B -@= 3-1 = (4+28) - (A+B) A = 1-2=-1 : Om From so the @ 1= i- A +ą solution of ® is - y = -2X+264x

1. © ya ley' +9y = 0, y Co) =2, 8(0) = -30 The auxiliary equation in this case is m?–12m +9 = 0 . i m = 12 + 5144 – 36 127 1108 . m = 12 + - J 144 3 2 - 2 6tat -27 2 = 12 = 2JR7 = 6 127 So the solution of @ is y = A + E) B (6-127) Be -> y'= (6+127) A estua +(6-127) Be Cova 3 700)=2 → 2 = A + y (0= -3 => -3 +(6+137) A + (6-vā7) B >© (14) + (6 + 027) - ③ we get 12 +2127423 (6+127) B - C6-177) 8 2465-272-8-7-3 ? 15+2127 = 23278 = B = 15+2 V 27, = It also ha = 1 + Putting the value of B in CW, we get the value of A. the solution is obtained from ® by putting the values of A&B. 2/27 Now