Author Topic: TUT0801 Quiz1  (Read 606 times)


  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
TUT0801 Quiz1
« on: September 27, 2019, 02:40:21 PM »
Q: t^3 y'+4t^2 y=e^(-t),t<0
    Find the general solution

  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
« Last Edit: September 27, 2019, 03:00:48 PM by XiaolongZhao »