## WeBWorK Problems

### Answer (in the form of a function) is not graded correctly

by Bentley Garrett -
Number of replies: 2
Hi:

We posted a problem with correct answer:

2/(t^2sqrt(8e^t-8e+1))

A student entered an incorrect answer:

1/(-(t^4)(8(e^t)-1-8e))^(1/2)

but it was graded as correct by WW.

This was the code that assigned the correct answer:

Context("Numeric");
Context()->variables->are(
t=>"Real"
);
parser::Assignment->Allow;

$answer = Formula(" 2/(t^2sqrt(8e^t-8e+1)) ")->with(limits=>[1,5]); The code for the comparison was: ANS($answer->cmp());

Can anyone tell what I did wrong?

Thanks!

### Re: Answer (in the form of a function) is not graded correctly

by Davide Cervone -
I am not able to reproduce the problem. The answer is marked incorrect for me. Can you provide the complete .pg file and the seed used in the problem?

### Re: Answer (in the form of a function) is not graded correctly

by Bentley Garrett -
Hi Davide:

The seed was: 176.

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

Also, if it means anything, we are using version 2.4 right now.

## 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();

"PGbasicmacros.pl",
"PGstandard.pl",
"MathObjects.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;