Solving (x+2y-4)dx - (2x-4y)dy = 0

[ronaldor9]

Homework Statement

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

The Attempt at a Solution

My answer,
$$\mathrm{ln}[4(y-1)^2+(x-2)^2] + 2\mathrm{arctan}\frac{x-2}{2y-2}=C$$
Textbook's answer,
$$\mathrm{ln}[4(y-1)^2+(x-2)^2] - 2\mathrm{arctan}\frac{2y-2}{x-2}=C$$

 

[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?

 

[ronaldor9]

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

 

[tiny-tim]

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)

 

[ronaldor9]

Thanks! tiny-tim