text: # Toronto Math Forum
  
  ## MAT244--2018F => MAT244--Tests => Quiz-1 => Topic started by: Victor Ivrii on September 28, 2018, 03:36:35 PM
  
  Title: Q1: TUT 0701
  Post by: Victor Ivrii on September 28, 2018, 03:36:35 PM
  Find the solution of the given initial value problem
  \begin{equation*}
  t^3y' + 4t^2y = e^{-t},\qquad  y(-1)=0,\qquad t<0.
  \end{equation*}
  Title: Re: Q1: TUT 0701
  Post by: Tzu-Ching Yen on September 28, 2018, 06:00:02 PM
  First divide by $t^3$ on both side of the equation, we get
  $$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$
  Using the method of integrating factor we have equation for $u(t)$
  $$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$
  where constant $c$ is arbitrary, it's chosen to be 0 here. Then
  $$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$
  rearranging gives equation
  $$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$
  substitute in $u(t) = t^4$
  $$y = \frac{1}{t^4}\int te^{-t}$$
  use integration by parts
  $$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$
  to check $c_1$, plug in condition $y(-1) = 0$
  $$y(-1) = e - e + c_1 = c_1= 0$$
  Plug in $c_1 = 0$ gets
  $$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4}$$
  Title: Re: Q1: TUT 0701
  Post by: Nick Callow on September 28, 2018, 06:00:48 PM
  My solution to this quiz can be found on this attachment.
  Title: Re: Q1: TUT 0701
  Post by: Victor Ivrii on September 29, 2018, 03:27:37 PM
  Nick, still typed in forum solution is better then typed externally (but I applaud your typesetting skills)