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\begin{document}

\begin{center}
{\Large Math 2233 \\[0pt]
Hint for HW 2 Problem 5}
\end{center}

Note: Since MyLab Math randomizes the homework problems for each student,
this version of problem 5 may not be exactly what MyLab Math assigned to you.

Problem 2.2.25 in text.

\begin{equation*}
x^{2}\frac{dy}{dx}=\frac{4x^{2}-x-2}{\left( x+1\right) \left( y+1\right) }%
\qquad ,\qquad y\left( 1\right) =1
\end{equation*}

\begin{itemize}
\item Mulitply both sides of the differential equation by $\left( y+1\right)
/x^{2}$ to get%
\begin{equation*}
\left( y+1\right) \frac{dy}{dx}=\frac{4x^{2}-x-2}{x^{2}\left( x+1\right) }
\end{equation*}%
or%
\begin{equation*}
-\frac{4x^{2}-x-2}{x^{2}\left( x+1\right) }+\left( y+1\right) \frac{dy}{dx}=0
\end{equation*}%
So the differential equation is of the separable form $M\left( x\right)
+N\left( y\right) \frac{dy}{dx}=0$, with%
\begin{equation*}
M\left( x\right) =-\frac{4x^{2}-x-2}{x^{2}\left( x+1\right) }\qquad ,\qquad
N\left( y\right) =y+1
\end{equation*}%
Therefore, we can solve it by solving instead%
\begin{equation*}
\int M\left( x\right) dx+\int N\left( y\right) dy=C
\end{equation*}%
Now
\begin{eqnarray*}
\int M\left( x\right) dx &=&-\int \frac{4x^{2}-x-2}{x^{2}\left( x+1\right) }%
dx \\
&=&-\frac{1}{x}\left( 3x\ln \left( x+1\right) +x\ln x+2\right) 
\end{eqnarray*}%
(This computation is done at the end of this hint.)
and%
\begin{equation*}
\int N\left( y\right) =\int \left( y+1\right) dy=\frac{1}{2}y^{2}+y
\end{equation*}%
Thus, we have%
\begin{equation}
-\frac{1}{x}\left( 3x\ln \left( x+1\right) +x\ln x+2\right) +\frac{1}{2}%
y^{2}+y=C  \tag{*}
\end{equation}

\item Now we're at a convenient point to determine the correct value for the
constant $C$. \ We apply the initial condition $y(1)=1$ by substitituting $%
x=1$, $y=1$ into (*). \ This yields%
\begin{equation*}
-\left( 2\ln \left\vert 2\right\vert +0+2\right) +\frac{1}{2}+1=C
\end{equation*}
\begin{equation*}
C=-\frac{1}{2}-2\ln \left\vert 2\right\vert
\end{equation*}

Thus, the implicit solution to the initial value problem is%
\begin{equation}
-\frac{1}{x}\left( 3x\ln \left( x+1\right) +x\ln x+2\right) +\frac{1}{2}%
y^{2}+y=-\frac{1}{2}-2\ln \left\vert 2\right\vert  \tag{**}
\end{equation}

\item Equation (**) is quadratic in $y$, and so it can be solved for $y$
using the Quadratic Formula%
\begin{equation*}
ay^{2}+by+c=0\quad \Rightarrow \quad y=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}
\end{equation*}%
with%
\begin{eqnarray*}
a &=&\frac{1}{2} \\
b &=&1 \\
c &=&\frac{1}{2}+2\ln \left\vert 2\right\vert -\frac{1}{x}\left( 3x\ln
\left( x+1\right) +x\ln x+2\right)
\end{eqnarray*}

So%
\begin{eqnarray*}
y\left( x\right) &=&-1\pm \sqrt{1-2\left( \frac{1}{2}+2\ln \left\vert
2\right\vert -\frac{1}{x}\left( 3x\ln \left( x+1\right) +x\ln x+2\right)
\right) } \\
&=&\sqrt{2}\sqrt{\frac{1}{x}\left( 3x\ln \left( x+1\right) -2x\ln 2+x\ln
x+2\right) }-1
\end{eqnarray*}

My apologies for such a messy computation. \ It may be that My Lab Math
expected you to the implicit solution (**) as your answer.
\end{itemize}


\bigskip
\begin{center}
Computation of $\int M(x) \, dx$
\end{center}


Note that the integrand of $\int M\left( x\right) dx$ can be written as%
\begin{equation*}
-\frac{4}{x+1}+\frac{1}{x\left( x+1\right) }+\frac{2}{x^{2}\left( x+1\right) 
}
\end{equation*}%
\par
We have%
\begin{equation*}
\int -\frac{4}{x+1}dx=-4\ln \left[ x+1\right] 
\end{equation*}%
\par
by substituting $u=x+1$. 
\par
\begin{equation*}
\int \frac{1}{x\left( x+1\right) }dx
\end{equation*}%
can be integrated using Partial Fractions%
\begin{eqnarray*}
\frac{1}{x\left( x+1\right) } &=&\frac{A}{x}+\frac{B}{x+1}\quad  \\
\quad  &\Rightarrow &\quad 1=A\left( x+1\right) +Bx \\
x &=&0\quad \Rightarrow \quad A=1 \\
x &=&-1\quad \Rightarrow \quad B=-1
\end{eqnarray*}%
and so%
\begin{eqnarray*}
\int \frac{1}{x\left( x+1\right) }dx &=&\int \frac{1}{x}dx-\int \frac{1}{x+1}%
dx \\
&=&\ln \left\vert x\right\vert -\ln \left\vert x+1\right\vert 
\end{eqnarray*}%
\par
\begin{equation*}
\int \frac{2}{x^{2}\left( x+1\right) }dx
\end{equation*}%
can also be done using Partial Fractions; however, the Partial Fraction
Expansion is slightly more complicated because of the $x^{2}$ term in the
denominator:%
\begin{equation*}
\frac{2}{x^{2}\left( x+1\right) }=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}
\end{equation*}%
\begin{eqnarray*}
\quad  &\Rightarrow &\quad 2=Ax\left( x+1\right) +B\left( x+1\right) +Cx^{2}
\\
x &=&0\quad \Rightarrow \quad B=2 \\
x &=&-1\quad \Rightarrow \quad C=2 \\
x &=&1\quad \Rightarrow \quad 2=2A+(2)(2)+(2)\quad  \\
&\Rightarrow &\quad A=-2
\end{eqnarray*}%
and so%
\begin{eqnarray*}
\int \frac{2}{x^{2}\left( x+1\right) }dx &=&\int \left( -\frac{2}{x}+\frac{2%
}{x^{2}}+\frac{2}{x+1}\right) dx \\
&=&-2\ln \left\vert x\right\vert -\frac{2}{x}+2\ln \left\vert x+1\right\vert 
\end{eqnarray*}%
Thus%
\begin{eqnarray*}
\int M\left( x\right) dx &=&-4\ln \left[ x+1\right]  \\
&&+\ln \left\vert x\right\vert -\ln \left\vert x+1\right\vert  \\
&&-2\ln \left\vert x\right\vert -\frac{2}{x}+2\ln \left\vert x+1\right\vert 
\\
&=&-\frac{1}{x}\left( 3x\ln \left( x+1\right) +x\ln x+2\right) 
\end{eqnarray*}%



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