Bonus problem for week 2

[Victor Ivrii]

Equation
$$
y'=\frac{y-x-2}{y+x}
$$
by a change of variables $x=t+a$, $y=z+b$reduce to homogeneous equation and solve it. Express $y$ as an implicit function of $x$:
$$
F(x,y)=C.
$$

[Changyu Li]

$
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
$$
\frac{1}{u} du = \frac{1+t}{-1-t^2}dt \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| t^2 +1 \right|  -\arctan t + C \\
\ln \left| u \right| = -\frac{1}{2}\ln \left| \left( \frac{v}{u} \right) ^2 +1 \right|  -\arctan \left( \frac{v}{u} \right) + C \\
\ln \left| x+1 \right| = -\frac{1}{2}\ln \left| \left( \frac{y-1}{x+1} \right) ^2 +1 \right|  -\arctan \left( \frac{y-1}{x+1} \right) + C
$$

[Victor Ivrii]

Please change a name to one which allows to identify you.

Correct final steps as $\int \frac{1}{1+t^2}\,dt$ calculated incorrectly.

Also type \tan \ln to produce $\tan, \ln$ etc; $\arctan(z)$ is preferable to $\tan^{-1}(z)$ which could be confused with $1/\tan(z)$.

Note: the final expression could be simplified.

[Changyu Li]

Changes were made per your suggestions.

[Victor Ivrii]

The final expression as I mentioned could be simplified
$$
\frac{1}{2}\ln \bigl( (y-1)^2+(x+1)^2 \bigr) + \arctan \left( \frac{y-1}{x+1} \right) = C.
$$

[Brian Bi]

let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$

What's the motivation for this? Inspired guess?

[Victor Ivrii]

let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$

What's the motivation for this? Inspired guess?

Standard method solving homogeneous equations i.e. of the form $y'=f(y/x)$.

[Brian Bi]

I see. Is this covered in class or in the textbook? I can't find it in the textbook.

[Changyu Li]

Problem 30 from chapter 2.2 in the tenth edition is a similar problem with a partial solution.