Differential Equation

ODE: t^3 x"- tx'+x=0, t>0 ---(A)
(a) x1(t)=t 代入(A),成立, 故x1(t)=t為(A)之一解
(b)(i) 設 x=ty,則 x'=y+ty', x"=2y'+ty", 代入(A)得
t^4 y"+(2t^3-t^2) y'=0, or
t^2 y"+(2t-1) y'=0 -----(B)
(ii) y"/y'= -2/t + 1/t^2
ln(y')= ln(1/t^2)- 1/t +a
y'=A/t^2 * exp(-1/t) + C
y= Aexp(-1/t) + C -----general sol. of (B)
(c) general sol. of (A)= ty= At exp(-1/t) + Ct
(d)
設(C) t^3 x"-tx'+x=t^3 之(特殊)解為x=u t exp(-1/t) + v t (u=u(t), v=v(t))
又設 u' t exp(-1/t)+ v' t=0 or
u' exp(-1/t) + v'=0 -------(D)
則
x'= u [1 + (1/t)] exp(-1/t) + v
x"= u 1/t^3 exp(-1/t)+ u' [1+(1/t)]exp(-1/t) + v'
代入(C)式,得: u' [1+(1/t)]exp(-1/t) + v' = 1 ----(E)
解(D),(E),得
u'= t exp(1/t)
v'= -t
積分(求一特殊解即可)得 v=-t^2/2, u=∫[1~t] xexp(1/x) dx=U(t)
特殊解x=t exp(-1/t)U(t) - t^3/2
(e) general sol. of (C) x(t)= t exp(-1/t) *[U(t)+A] +Ct - t^3/2 )
x'(t)=(1 + 1/t) exp(-1/t)*[U(t)+A]+ t^2 + C - 3t^2/2
x(1)=0= 1/e *[0+A]+C - 1/2 =0
x'(1)=1= 2/e*[0 +A]+1+C- 3/2=1, 則A=e, C=-1/2
故x(t)= t exp(-1/t)[U(t)+ e] - (t+ t^3)/2, 其中U(t)=∫[1~t] x exp(1/x) dx