solve dy/dx = (1+y^2)/(1+x^2)

[doive]

I'm picking up my maths again after a number of years of rust and I'm having some trouble.

Obtain the solution to the DE:
dy/dx = (1+y²) / (1+x²)

My solution:
dy / (1+y²) = dx / (1+x²)
knowing that dx/(1+x²) = arctan(x) (I believe this is a standard result? I've followed some trig to make myself comfortable with it in any case.)
arctan(y) = arctan(x) + A
y = x + C

The question I'm trying to solve is multiple choice and all the options I have are mixed fractions such as Cx/(1-Cx) or (1-Cx) / (x+C).
Intuitively it seems like y = x + C isn't an appropriate solution as that leads to dy/dx = 1 which seems contradictory.
Where am I going wrong?!

 

[Ishuda]

I'm picking up my maths again after a number of years of rust and I'm having some trouble.

Obtain the solution to the DE:
dy/dx = (1+y^2) / (1+x^2)

My solution:
dy / (1+y^2) = dx / (1+x^2)
knowing that dx/(1+x^2) = arctan(x) (I believe this is a standard result? I've followed some trig to make myself comfortable with it in any case.)
arctan(y) = arctan(x) + A
y = x + C

The question I'm trying to solve is multiple choice and all the options I have are mixed fractions such as Cx/(1-Cx) or (1-Cx) / (x+C).
Intuitively it seems like y = x + C isn't an appropriate solution as that leads to dy/dx = 1 which seems contradictory.
Where am I going wrong?!

If you have the solution
arctan(y) = arctan(x) + A
and take the tangent across you get
tan(arctan(y)) = y = tan( arctan(x) + A )
What is the formula for the sum of angles for the tangent?

 

[doive]

If you have the solution
arctan(y) = arctan(x) + A
and take the tangent across you get
tan(arctan(y)) = y = tan( arctan(x) + A )
What is the formula for the sum of angles for the tangent?

That's where I'm going wrong - Thanks so much!
The rust clearly has gone deeper than I thought

For completeness:
tan(A+B) = (tan(A)+tan(B))/(1-tan(A)*tan(B))
with A = arctan(x)
y = (x + tan(B))/(1-x*tan(B)
meaning my solution is: y = (x + C)/(1-Cx)

Thanks Ishuda!