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Ch 1--2 / Re: Bonus problem for week 2
« on: January 18, 2013, 01:41:07 AM »
$
x = u+h \\
y = v+k\\
$
at $(u,v) = 0\\$
$
k - h - 2 = 0 \\
h + k = 0 \\
\Rightarrow h = -1,\;k = 1\\
$
$
x = u - 1 \\
dx = du\\
y = v + 1 \\
dy = dv \\
$
$$
\frac{dv}{du} = \frac{v-u}{u+v} \\
$$
let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$
$$
t + u \frac{dt}{du}=\frac{ut-u}{u+ut} = \frac{t-1}{1+t}
$$
simplify with magic
$$
\frac{1}{u} du = \frac{1+t}{-1-t^2}dt \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| t^2 +1 \right| -\arctan t + C \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| \left( \frac{v}{u} \right) ^2 +1 \right| -\arctan \left( \frac{v}{u} \right) + C \\
\ln \left| x+1 \right| = -\frac{1}{2}\ln \left| \left( \frac{y-1}{x+1} \right) ^2 +1 \right| -\arctan \left( \frac{y-1}{x+1} \right) + C
$$
x = u+h \\
y = v+k\\
$
at $(u,v) = 0\\$
$
k - h - 2 = 0 \\
h + k = 0 \\
\Rightarrow h = -1,\;k = 1\\
$
$
x = u - 1 \\
dx = du\\
y = v + 1 \\
dy = dv \\
$
$$
\frac{dv}{du} = \frac{v-u}{u+v} \\
$$
let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$
$$
t + u \frac{dt}{du}=\frac{ut-u}{u+ut} = \frac{t-1}{1+t}
$$
simplify with magic
$$
\frac{1}{u} du = \frac{1+t}{-1-t^2}dt \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| t^2 +1 \right| -\arctan t + C \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| \left( \frac{v}{u} \right) ^2 +1 \right| -\arctan \left( \frac{v}{u} \right) + C \\
\ln \left| x+1 \right| = -\frac{1}{2}\ln \left| \left( \frac{y-1}{x+1} \right) ^2 +1 \right| -\arctan \left( \frac{y-1}{x+1} \right) + C
$$