Q: ty’ +(t+1)y=t. y(ln(2))=1
Ans:
y’+[(t+1)/t]y=1
p(t)=[(t+1)/t]
u=e^(∫[(t+1)/t]dt)=te^t
te^t*y’+[(t+1)/t]*te^t*y=te^t
te^t *y=∫te^tdt
u=t du=1dt
v=e^t dv=e^tdt
te^t *y = te^t-∫e^t dt = te^t - e^t +c
y=1-(1/t)+(c/(te^t))
y=1 t=ln2
1=1-1/ln2 +c/(ln2)
C=2
y=1-1/t+2/(te^t)
