MAT244-2013S > Ch 1--2

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.

$$

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