\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
\parindent=0pt
\begin{document}

{\bf Question}

Find two different solutions (one can come from the general
solution and the other from a singular solution) to the equation
$$ \left(\dy\right)^2 - x\dy +y =0 $$ which satisfy the condition
$y(1)=1/4. \qquad (*)$



\vspace{0.25in}

{\bf Answer}

First rewrite in standard form. $\ds
y=x\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^2$, now differentiate
to get

$\ds\frac{dy}{dx}=\frac{dy}{dx}+x\frac{d^2y}{dx^2}-2\frac{dy}{dx}\frac{d^2y}{dx^2}
\Rightarrow 0=\frac{d^2y}{dx^2}\left(x-2\frac{dy}{dx}\right)$

hence
\begin{itemize}
\item[i)] $\ds\frac{d^2y}{dx^2}=0$ and $\ds\hspace{0.1in}
y=x\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^2$

$\ds\hspace{0.1in} \Downarrow \hspace{1.4in} \Downarrow$

$\ds\frac{dy}{dx}=c \hspace{0.15in} \Rightarrow \hspace{0.15in}
y=cx-c^2 \hspace{0.5in}$ this is the general solution.

\item[ii)] $\ds x-2\frac{dy}{dx}=0$ and
$\ds y=x\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^2$

eliminate $\ds\frac{dy}{dx} \Rightarrow y=\frac{x^2}{4}$  this is
a singular solution.

$\ds \begin{array}{ll} y(1)=\frac{1}{4} & \Rightarrow {\rm
singular\ solution\ } y=\frac{x^2}{4} {\rm is\ valid}.\\ &
\Rightarrow {\rm general\ solution\ } y=cx-c^2 {\rm with\ }
\frac{1}{4}=A-A^2\end{array}$

$\ds \hspace{1.5in} \Rightarrow A=\frac{1}{2}$ hence $\ds
y=\frac{x}{2}-\frac{1}{4}$ is a solution.

\end{itemize}


\end{document}