The seed was: 176.

Below is the complete code for the .pg file. Sorry about the length, but I included everything to be safe.

## DESCRIPTION

## First order ODEs: Bernoulli

## ENDDESCRIPTION

## KEYWORDS('differential equations','first order','Bernoulli')

## DBsubject('Differential Equations')

## DBchapter('First Order Differential Equations')

## DBsection('Bernoulli')

## Date('10/10/2012')

## Author('ZH')

## Institution('UTD')

## TitleText1('')

## EditionText1('')

## AuthorText1('')

## Section1('')

## Problem1('')

##############################

# Initialization

DOCUMENT();

loadMacros(

"PGbasicmacros.pl",

"PGstandard.pl",

"MathObjects.pl",

"AnswerFormatHelp.pl",

"parserAssignment.pl",

);

TEXT(beginproblem());

#############################

# Setup

Context("Numeric");

Context()->variables->are(

t=>"Real"

);

parser::Assignment->Allow;

$answer = Formula(" 2/(t^2sqrt(8e^t-8e+1)) ")->with(limits=>[1,5]);

######################

Context()->texStrings;

BEGIN_TEXT

Solve the following Initial Value Problem

$BR

$BR

\( \begin{cases} ty' + 2y + t^5y^3e^t = 0 \\ y(1)=2 \end{cases} \)

$BR

$BR

\( y=\) \{ ans_rule(60) \}

\{ AnswerFormatHelp("formulas") \}

END_TEXT

Context()->normalStrings;

Context()->texStrings;

SOLUTION(EV3(<<'END_SOLUTION'));

$PAR SOLUTION $PAR

This is a Bernoulli equation

$BR

$BR

\[ y' + \frac2t y = -t^4y^3e^t \]

$BR

$BR

Thus, we use the substitution \( u = y^{1-3} = y^{-2} \), i.e. \( y = u^{-\frac12} \) and \( y' = -\frac12 u^{-\frac32}u' \) and we get

$BR

$BR

\[ -\frac12u^{-\frac32}u' + \frac2t u^{-\frac12} = -t^4u^{-\frac32}e^t \iff u' - \frac4tu = 2t^4e^t \]

$BR

$BR

The last equation is linear, therefore we use the integrating factor \( \mu = e^{-\int \frac{4dt}t} = \frac1{t^4} \) and we get

$BR

$BR

\[

\frac d{dt}\left(\frac u{t^4}\right)=2e^t\;\; \Leftrightarrow\;\; y^{-2}=u=2t ^4e^t+Ct^4

\]

$BR

$BR

And we obtain the following general solution:

$BR

$BR

\[ y = \frac{\pm 1}{\sqrt{2t^4e^t+Ct^4}}, \text{ and } y \equiv 0 \]

$BR

$BR

The initial value gives us:

$BR

$BR

\[ 2 = \frac{1}{\sqrt{2e+C}} \Rightarrow 2e+C = \frac14 \Rightarrow C = \frac{1-8e}{4} \]

$BR

$BR

which leads to the solution

$BR

$BR

\[ y = \frac2{t^2\sqrt{8e^t-8e+1}} \]

END_SOLUTION

Context()->normalStrings;

######################

# Answer evaluation

$showPartialCorrectAnswers = 1;

ANS( $answer->cmp( ));

COMMENT("MathObject version.");

ENDDOCUMENT();