1. ## Differential equation

Hi I am trying to solve the following differential equation but not having much luck!

dy/dx = 1+y^2 / 1 + x^2

Its a multiple choice question and the answer is one of the following:

(A) Cx / (1 - Cx)

(B) Cx / (1 + Cx)

(c) (C - x) / (1 - Cx)

(D) (1 - Cx) / (x + C)

(E) (x + C) / (1 - Cx)

Here is what I have so far:

Rearange to the form:

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

Integrate both sides with respect to x, giving:

tan^-1(y) = tan^-1(x) + c

which is the same as:

y = tan(tan^-1(x) + c)

y = x + tan(c)

we can let C = tan(c) as it is constant, giving:

y = x + C

but this looks nothing like the answer.

Any help much appreciated, have spent ages trying to solve this!

2. Originally Posted by alexthedon
y = tan(tan^-1(x) + c)

y = x + tan(c)
Here you messed up.

3. ## I thought that...

I thought that tan(tan^-1(x)) = x ?

am I wrong there is is it something to do with the c?

4. Yeah, but just works for that case.

What about if you apply the formula $\tan(\alpha+\beta)$?

5. ## Ahhh I see

Many thanks.

y = \tan(tan^-1(x) + c)

Using:

$
\tan(\alpha+\beta)
$

we get:

tan(tan^-1(x)) + tan(c) / 1 - tan(tan^-1(x)) * tan(c)

which gives us:

x + tan(c) / 1 + x.tan(c)

if we now let C = tan(c)

we have:

x + c / 1 - Cx

I obviously need to know my Trig Cheat sheet better!

Cheers,

Alex