Q: t^3 y'+4t^2 y=e^(-t),t<0
y(-1)=0
Find the general solution
A:
Step1: let the coefficient of y' be '1'
y'+4/t y=e^(-t)/t^3
Step2: Set u = e^∫〖p(t) dt〗, where p(t) is the coefficient of y
Set u=e^∫〖4/t dt〗=e^(4ln|t|))=e^(4ln(-t))=e^(ln((-t)^4))=e^(ln(t^4))=t^4 (since t<0)
Step3: Multiply u to both sides
t^4 y'+4t^3 y=te^(-t)
(t^4 y)'=te^(-t)
Step4: Set integral to both sides
∫〖(t^4 y)' dt〗=∫〖te^(-t) dt〗
For ∫〖te^(-t) dt〗, use ‘by parts’ to solve it:
u=t, du=1*dt,
dv=e^(-t)dt, v= -e^(-t)
∫〖te^(-t) dt〗=-te^(-t)-∫〖-e^(-t) dt〗=-te^(-t)-e^(-t)+C
t^4 y= -te^(-t)-e^(-t)+C
y= -e^(-t)/t^3 -e^(-t)/t^4 +C/t^4
Step5: Plug in y(-1)=0
0=e-e+C, thus, C=0
Thus, y= -e^(-t)/t^3 -e^(-t)/t^4
