(a)
y=t*u
y'=u+tu'
y''=u'+u'+tu''=tu''+2u'
代入方程
t^2(tu''+2u')-t(t+2)(tu'+u)+(t+2)tu=0
t^3u''+(2t^2-t^2(t+2))u'=0
u''+[(-1-t)/t]u'=0
令v=u'
v'+[(-1-t)/t]v=0
integrating factor z=exp(int P(t)dt)=exp(-t-lnt)=e^(-t)/t
apply z
(vz)'=0
vz=C
v=C/z=Cte^t
u=int v dt =C(t-1)e^t
y=Ct(t-1)e^t
Combine y1 and y2, we have the general solution
C1*t+C2*t(t-1)e^t
(b)
y=u*e^t
y'=u'e^t+ue^t
y''=u''e^t+2u'e^t+ue^t
t(u''e^t+2u'e^t+ue^t)-(t+2)(u'e^t+ue^t)+2ue^t=0
te^t u''+(t-2)e^t u'=0
v=u'
v'+[(t-2)/t] v=0
z=exp(int (t-2)/t dt)=exp(t-2lnt)=e^t/t^2
(vz)'=0
vz=C
v=C/z=Ct^2e^(-t)
u=int v dt=Ce^(-t)(-t^2-2t-2)
y=C1e^t+C2e^(-t)(-t^2-2t-2)
(c)
y=u t^(-1/2)sint
y'=u't^(-1/2)sint+u[(-1/2)t^(-3/2)sint+t^(-1/2)cost]
y''=u''t^(-1/2)sint+2u'[(-1/2)t^(-3/2)sint+t^(-1/2)cost]+u[(3/4)t^(-5/2)sint-t^(-3/2)cost-t^(-1/2)sint]
带入
u''t^(3/2)sint+u'[2t^(3/2)cost]=0
v=u'
v'+2cot t v=0
dv/dt=-2cot t v
dv/v=-2cot t dt
ln v=-2ln(sint)
v=(sint)^(-2)=csc^2 t
u=int v dt =- cot t
y=-t^(-1/2) cost
所以
y=C1t^(-1/2) cost+C2t^(-1/2) sint
