MAT244--2018F > Quiz-1

Q1: TUT 0701

(1/1)

**Victor Ivrii**:

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**:

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

$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} $$

**Nick Callow**:

My solution to this quiz can be found on this attachment.

**Victor Ivrii**:

Nick, still typed in forum solution is better then typed externally (but I applaud your typesetting skills)

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