# Homogenous ode

1. Sep 11, 2009

### Mechdude

1. The problem statement, all variables and given/known data
the homogenous ode
$$(x^2+y^2)dy-y^2 dx =0$$

2. Relevant equations

$v=\frac{y}{x}$

3. The attempt at a solution
worked al the way here $$\int \frac {1}{v^2-v(1+v^2) }dv + \int\frac{v^2}{v^2- v(1+v^2)} = \int \frac {dx}{x}$$
how do i handle the integrals ?

2. Sep 11, 2009

### HallsofIvy

Staff Emeritus
I would first put it back together again:
$$\int\frac}(1+ v^2)dv}{v^2- v- v^3}= -\int\frac{(1+ v^2)dv}{v(v^2- v+ 1)}$$
and, since it is a rational integral, use partial fractions.

3. Sep 12, 2009

### Mechdude

The integral i get after getting partial fractions has me beat: i got A=1, B=-1
$$\int\frac {1}{-v}dv - \int\frac {1}{v^2-v+1} =\int \frac{dx}{x}$$
the second integral on the left is quadratic and irriducible how is it integrated?

4. Sep 12, 2009

### Mechdude

Im thinking about completing the square thus the second integral on the left becomes:
$$- \int \frac{dv} {(v-\frac{1}{2} )^2 + \frac{3}{4} }$$

Last edited: Sep 12, 2009
5. Sep 12, 2009

### HallsofIvy

Staff Emeritus
Yes, that's correct. An irreducible quadratic term can always be written as $(v- a)^2+ b$ and then the substitution $u= (v- a)/\sqrt{b}$ reduces it to $(1/b)(1/(u^2+ 1))$ and the integral is an arctangent.

Last edited: Sep 12, 2009
6. Sep 12, 2009

Thanks