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

## Homework Statement

$$(x+2y-4)\,dx - (2x-4y)\,dy=0$$

## The Attempt at a Solution

$$\mathrm{ln}[4(y-1)^2+(x-2)^2] + 2\mathrm{arctan}\frac{x-2}{2y-2}=C$$
$$\mathrm{ln}[4(y-1)^2+(x-2)^2] - 2\mathrm{arctan}\frac{2y-2}{x-2}=C$$

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tiny-tim
Homework Helper
Hi ronaldor9!

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

what were the initial conditions?

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

tiny-tim
Homework Helper
Hey, sorry i meant to place a constant instead of the zero.
Does that identity work for arctan?
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)

Thanks! tiny-tim