# Homework Help: Can anyone check if my answer for this ODE is correct?

1. Dec 23, 2009

### ronaldor9

1. The problem statement, all variables and given/known data
$$(x+2y-4)\,dx - (2x-4y)\,dy=0$$

3. The attempt at a solution
$$\mathrm{ln}[4(y-1)^2+(x-2)^2] + 2\mathrm{arctan}\frac{x-2}{2y-2}=C$$
$$\mathrm{ln}[4(y-1)^2+(x-2)^2] - 2\mathrm{arctan}\frac{2y-2}{x-2}=C$$

Last edited: Dec 23, 2009
2. Dec 23, 2009

### tiny-tim

Hi ronaldor9!

tan(a) = 1/tan(π/2 - a), so theyr'e the same, except for a constant

what were the initial conditions?

3. Dec 23, 2009

### ronaldor9

Hey, sorry i meant to place a constant instead of the zero.
Does that identity work for arctan?

4. Dec 23, 2009

### tiny-tim

oh come on

if tan(a) = 1/tan(π/2 - a),

then putting T = tan(a) gives 1/T = tan(π/2 - a),

so arctanT = a = π/2 - arctan(1/T)

5. Dec 23, 2009

### ronaldor9

Thanks! tiny-tim