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\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}

{\bf Question}

If $ad-bc \neq 0$, show that constants $h$ and $k$ can be chosen
in such a way that the change of variables $X=x+h$, $Y=y+k$
reduces

\[
\dy={{ax+by +p} \over {cx+dy+q}}
\]

to a homogeneous equation.

Hence,or otherwise,  solve the equation

\[
\dy={{x+y+4} \over {x-y-6}}
\]




\vspace{0.25in}

{\bf Answer}

The ODE becomes:
$\ds\frac{dY}{dX}=\frac{a(X-h)+b(Y-k)+p}{c(X-h)+d(Y-k)+q}$

$\ds\frac{dY}{dX}=\frac{aX+bY+(p-ah-bk)}{cX+dY+(q-ch-dk)}$ for
homogeneous we need

$\ds\left.\begin{array}{l} p-ah-bk=0\\
q-ch-dk=0\end{array}\right\}$ which has a unique solution if $\ds
ad-bc\not=0$

and the equation then becomes
$\ds\frac{dY}{dX}=\frac{a+b\frac{Y}{X}}{c+d\frac{Y}{X}}$

For example if $\ds a=1,\,\,\,\, b=1,\,\,\,\, p=4,\,\,\,\,
c=1,\,\,\,\, d=-1,\,\,\,\, q=-6$

choose $\ds h,\,\, k$ so that $\ds\left.\begin{array}{l} h+k=4\\
h-k=-6\end{array}\right\} \Rightarrow h=-1,\,\,\,\, k=5$

$\ds X=x-1,\,\,\,\, Y=y+5$ and the ODE becomes

$\ds\frac{dY}{dX}=\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$ \,\,\, put
$\ds Y=Xv \Rightarrow X\frac{dv}{dX}+v=\frac{1+v}{1-v}$


$\ds X\frac{dv}{dX}=\frac{1+v^2}{1-v} \Rightarrow
\int\frac{1-v}{1+v^2}dv=\int\frac{1}{X}dX$

$\ds\Rightarrow \int\frac{1}{1+v^2}dv-\int\frac{v}{1+v^2}dv=\ln
X+c$

$\ds \Rightarrow \tan^{-1}v-\frac{1}{2}\ln(1+v^2)=\ln X+c$

or $\ds \tan^{-1}\frac{Y}{X}=\ln(X^2+Y^2)^\frac{1}{2}+c$

$\ds \Rightarrow
\tan^{-1}\left(\frac{y+5}{x-1}\right)=\ln(x^2+y^2-2x+10y+26)^\frac{1}{2}+c$



\end{document}
