Toronto Math Forum
MAT2442013S => MAT244 MathLectures => Ch 12 => Topic started by: Victor Ivrii on January 17, 2013, 02:18:45 PM

Equation
$$
y'=\frac{yx2}{y+x}
$$
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 $x$:
$$
F(x,y)=C.
$$

$
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 \\
$
$$
\frac{dv}{du} = \frac{vu}{u+v} \\
$$
let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$
$$
t + u \frac{dt}{du}=\frac{utu}{u+ut} = \frac{t1}{1+t}
$$
simplify with magic
$$
\frac{1}{u} du = \frac{1+t}{1t^2}dt \\
\ln \left u \right = \frac{1}{2}\ln \left t^2 +1 \right \arctan t + C \\
\ln \left u \right = \frac{1}{2}\ln \left \left( \frac{v}{u} \right) ^2 +1 \right \arctan \left( \frac{v}{u} \right) + C \\
\ln \left x+1 \right = \frac{1}{2}\ln \left \left( \frac{y1}{x+1} \right) ^2 +1 \right \arctan \left( \frac{y1}{x+1} \right) + C
$$

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.

Changes were made per your suggestions.

The final expression as I mentioned could be simplified
$$
\frac{1}{2}\ln \bigl( (y1)^2+(x+1)^2 \bigr) + \arctan \left( \frac{y1}{x+1} \right) = C.
$$

let $v = ut,\;\frac{dv}{du}=t+u \frac{dt}{du}$
What's the motivation for this? Inspired guess?

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 $y'=f(y/x)$.

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

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