Trouble Indenting Portion of document

Question

This is my first time giving LaTeX a try, and I decided to have my go at it with my week's math homework. Though most of the document could probably be done in a more efficient manner, I've gone about 30 lines without too much of a hitch.

My problem arises in line 30, where I wanted to format a line so that I had text flushed left and right on the same line, which I was able to do after some googling. However, the paragraphs following my \raggedleft text will not indent, no matter what I try. Here is the code, can somebody help me out?

    \documentclass[11pt]{article}
   \usepackage{framed}
   \usepackage{fullpage}
   \usepackage{amsmath}
   \author{Author Name Here}
   \title{Quiz 7 Makeup}
   \date{March 11, 2014}

   \begin{document}
   % generates the title
   \maketitle
  Instructions: Find general solutions of the differential equations for the following problems found in section 1.6:\\
  2) $xy^2+3y^2 -x^2y'=0$

  \underline{Solution:} $xy^2+3y^2 -x^2y'=0 \Rightarrow xy^2+3y^2=x^2y' \equiv y^2(3+x)=x^2y'$

  $LHS=\dfrac{x+3}{x^2}=\dfrac{y'}{y^2}=RHS$

  $LHS= \displaystyle{\int \frac{x+3}{x^2}}\,dx\ = \int \displaystyle{\frac{1}{x}+\frac{3}{x^2} \,dx} = \ln(x) +-3x^{-1} $

  $RHS= \displaystyle{\int \frac{y'}{y^2} \,dx = -y^{-1}}$

  Since $LHS=RHS$, we find the general solution to be $$y=\dfrac{1}{3x^{-1}-\ln{x}} .$$      \\
  8) $x^{2}y'=xy+x^{2}e^{y/x}$                        

  \underline{Solution:}  $x^{2}y'=xy+x^{2}e^{y/x} \equiv y'= \dfrac{y}{x}+e^{y/x} \Rightarrow y'-\dfrac{y}{x}=e^{y/x}$

  \indent\indent Integrating Factor: $e^{\int \frac{-1}{x}\,dx} = e^{-\ln{x}}=e^{\ln{x^{-1}}}=\dfrac{1}{x}$

  \indent$\frac{d}{dy}[y(x)]=e^{y/x}(\dfrac{1}{x}) \Rightarrow yx = \int \dfrac{1}{x}e^{y/x}\,dy = $  \hfill\raggedleft\framebox{Let $u =\dfrac{y}{x}$ so that $du= \dfrac{1}{x} \,dy$} \raggedright \flushleft
  \indent What works?

  What works?\\
  \indent What works?!

  \indent What works?
   \end{document}

As a side question, I would appreciate any tips if ya'll see some inefficiencies in my document. I'm particularly wondering about the sections below the math problems where I would like to have them indented; my solution was to make each new line a new paragraph, giving it another indentation. Is there a command I could use to to just indent a whole block of text?

Thank you! I greatly appreciate any help and advice.

Answer 1

I fixed some of the main issues, although I'm not sure of the layout you wanted.

\raggedright affects paragraphs at a time, and stays in force until the end of the group, or end of the document, so you didn't want that there. Similarly \displaystyle is a (rarely used) declaration that affects the entire math list, it does not take an {} argument. It is better to use a display math environment (\[ \] or alignetc). I used a list environment for your questions which sets the paragraph indentation to 0 so I used the enumitem package to set it back to something larger. Don't use the math italic font for multi-letter identifiers, so I switched it \mathit (which is text italic, in math mode). Try to avoid using \\ outside its use in alignments.

  \documentclass[11pt]{article}
   \usepackage{framed}
   \usepackage{fullpage}
   \usepackage{amsmath,enumitem}
   \author{Author Name Here}
   \title{Quiz 7 Makeup}
   \date{March 11, 2014}

   \begin{document}
   % generates the title
   \maketitle
  Instructions: Find general solutions of the differential equations for the following problems found in section 1.6:

\begin{enumerate}[listparindent=1cm]
 \item[2] $xy^2+3y^2 -x^2y'=0$

  \textbf{Solution:}
  \begin{gather*}
   xy^2+3y^2 -x^2y'=0 \Rightarrow xy^2+3y^2=x^2y' \equiv y^2(3+x)=x^2y'\\
  \mathit{LHS}=\frac{x+3}{x^2}=\frac{y'}{y^2}=\mathit{RHS}\\
  \mathit{LHS}= \int \frac{x+3}{x^2}\,dx = \int \frac{1}{x}+\frac{3}{x^2} \,dx = \ln(x) +-3x^{-1} \\
  \mathit{RHS}= \int \frac{y'}{y^2} \,dx = -y^{-1}
\end{gather*}

  Since $\mathit{LHS}=\mathit{RHS}$, we find the general solution to be
  \[y=\frac{1}{3x^{-1}-\ln{x}} .\]

  \item[8] $x^{2}y'=xy+x^{2}e^{y/x}$                        

  \textbf{Solution:} 
\[x^{2}y'=xy+x^{2}e^{y/x} \equiv y'= \dfrac{y}{x}+e^{y/x} \Rightarrow y'-\dfrac{y}{x}=e^{y/x}\]

  Integrating Factor:
\[e^{\int \frac{-1}{x}\,dx} = e^{-\ln{x}}=e^{\ln{x^{-1}}}=\dfrac{1}{x}\]

  $\frac{d}{dy}[y(x)]=e^{y/x}(\dfrac{1}{x}) \Rightarrow yx = \int \dfrac{1}{x}e^{y/x}\,dy = $ 
 \hfill 
 \framebox{Let $u =\dfrac{y}{x}$ so that $du= \dfrac{1}{x} \,dy$}


What works?

  What works?   What works?! this is an indent at start of para adding more text so it shows.
this is an indent at start of para adding more text so it shows.


  What works?
\end{enumerate}
   \end{document}