Author Topic: Q1: TUT 0701  (Read 1404 times)

Victor Ivrii

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Q1: TUT 0701
« 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*}

Tzu-Ching Yen

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Re: Q1: TUT 0701
« Reply #1 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
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$
« Last Edit: September 29, 2018, 03:24:11 PM by Victor Ivrii »

Nick Callow

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Re: Q1: TUT 0701
« Reply #2 on: September 28, 2018, 06:00:48 PM »
My solution to this quiz can be found on this attachment.

Victor Ivrii

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Re: Q1: TUT 0701
« Reply #3 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)