\documentclass[a4paper,12pt]{article}
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\begin{document}


{\bf Question}

Solve the following first order linear differential equations:
\begin{description}
\item[(a)] $xy'+y=\sin x$
\item[(b)]$(1-x^2)y'+xy=x$
\item[(c)] $y'=(y-1)\cot x$
\item[(d)] $x^3y'+(2-3x^2)y=x^3$
\end{description}



\vspace{.5in}

{\bf Answer}

\begin{description}
\item[(a)]
$\ds xy' + y = \sin x \hspace{.2in} (1) \hspace{.2in} \Rightarrow
\hspace{.2in} y' + \frac{y}{x} = \frac{\sin x}{x} \hspace{.2in}
(2) $

Integrating Factor $\ds e^{\int \frac{1}{x} dx} = e^{(ln x)} = x$

So (2) becomes \begin{eqnarray*} xy' + y & =  & \sin x \\
\Rightarrow \frac{d}{x}(xy) & = & \sin x \\ \Rightarrow xy &  = &
\int \sin \, dx \\  & = & -\cos x + c \\ y & = & -\frac{\cos x}{x}
+ \frac{c}{x} \end{eqnarray*}

\item[(b)]
$\ds (1 - x^2) y' + xy = x \hspace{.2in} (1) \hspace{.2in}
\Rightarrow \hspace{.2in} y' + \frac{x}{1 - x^2}y = \frac{x}{1 -
x^2} \hspace{.2in} (2)$

Integrating Factor $\ds e^{\int \frac{x}{1 - x^2} dx} =
e^{(-\frac{1}{2} \ln (1 - x^2))} = (1 - x^2)^{-\frac{1}{2}}$

so (2) becomes \begin{eqnarray*} \frac{y'}{(1 -
x^2)^{\frac{1}{2}}} + \frac{x}{(1 - x^2)^{\frac{3}{2}}}y & = &
\frac{x}{(1 - x^2)^{\frac{3}{2}}} \\ \Rightarrow \frac{d}{dx}
\left( \frac{y}{(1 - x^2)^{\frac{1}{2}}}\right) & = & \frac{x}{(1
- x^2)^{\frac{3}{2}}} \\ \Rightarrow \frac{y}{(1 -
x^2)^{\frac{1}{2}}} & = & \int \frac{x}{(1 - x^2)^{\frac{3}{2}}}
dx \\ & = & (1 - x^2)^{-\frac{1}{2}} + c \\ \Rightarrow y & = & 1
+ c\sqrt{1 - x^2}\end{eqnarray*}

\newpage
\item[(c)]
$\ds y' = (y - 1)\cot x \hspace{.2in} (1) \hspace{.2in}
\Rightarrow \hspace{.2in} y' - \cot (x) y = -\cot x \hspace{.2in}
(2)$

Integrating factor $ \ds e^{\int -\cot x \, dx} = e^{-\ln (\sin
x)} = (\sin x )^{-1}$

so (2) becomes \begin{eqnarray*} \frac{y'}{\sin x} + \frac{\cos
x}{\sin^2 x} y & = & -\frac{\cos x}{\sin^2x} \\ \Rightarrow
\frac{d}{dx} \left( \frac{y}{\sin x} \right) & = & -\frac{\cos
x}{\sin^2x} \\ \Rightarrow \frac{y}{\sin x} & = & -\int \frac{\cos
x}{\sin^2x} dx \\ & = & \frac{1}{\sin x} + c \\ \Rightarrow y & =
& 1 + c \sin x \end{eqnarray*}


\item[(d)]
$\ds x^3y' + (2 - 3x^2)y = x^3 \hspace{.2in} (1) \hspace{.2in}
\Rightarrow \hspace{.2in} y' + \left(  \frac{2}{x^3} -
\frac{3}{x}\right)y = 1 \hspace{.2in} (2)$

Integrating factor $\ds e^{\left(  \frac{2}{x^3} -
\frac{3}{x}\right) \, dx} = e^{-\frac{1}{x^2} - 3 \ln x} =
e^{-\frac{1}{x}} e^{-3 \ln x} = \frac{1}{x^3}e^{-\frac{1}{x^2}}$

so (2) becomes \begin{eqnarray*} \frac{y'}{x^3}e^{-\frac{1}{x^2}}
+ \left( \frac{2}{x^6} - \frac{3}{x^4}\right) e^{-\frac{1}{x^2}}y
& = & \frac{1}{x^3}e^{-\frac{1}{x^2}} \\ \Rightarrow \frac{d}{dx}
\left( \frac{y}{x^3}e^{-\frac{1}{x^2}}\right) & = & \frac{1}{x^3}
e^{-\frac{1}{x^2}} \\ \Rightarrow \frac{y}{x^3} e^{-\frac{1}{x^2}
} & = & \int \frac{1}{x^3} e^{-\frac{1}{x^2}} \, dx \\ & = &
\frac{1}{2}e^{-\frac{1}{x^2}} + c \\ \Rightarrow y & = &
\frac{x^3}{2} + cx^3 e^{\frac{1}{x^2}} \end{eqnarray*}


\end{description}



\end{document}
