Solution to differential equation

[TedMurphy]

Obtain the solution to the differential equation:

$$\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$$

Multiple choice answer:

a) $$\frac{Cx}{1-Cx}$$
b) $$\frac{Cx}{1+Cx}$$
c) $$\frac{C-x}{1-Cx}$$
d) $$\frac{1-Cx}{x+C}$$
e) $$\frac{x+C}{1-Cx}$$

Tried integrating two sides to arrive at arctan y = arctan x + C, but not sure how to proceed from there.

 

[CAF123]

Rearrange first to get $$\tan^{-1}y - \tan^{-1}x = C$$ Then take tan on both sides and simplify.

 

[TedMurphy]

I'm going to write the simplification out, because it took me a while.

$$\tan^{-1}y - \tan^{-1}x = C$$

In order to take tan on the left side, this equation needs to be re-written as:

$$\tan^{-1}\frac{y-x}{1+yx}=C$$

then we can take the tangent of both sides, giving us:

$$\frac{y-x}{1+yx}=C$$

then we solve for y:

$${y-x}={C+Cyx}$$

$$0 = C+Cyx-y+x$$

$$0 = y(Cx-1)+C+x$$

$$y = \frac{-C-x}{Cx-1}$$

$$y = \frac{x+C}{1-Cx}$$

Answer E above

 

[CAF123]

I agree with the answer E, but you can take tan from the beginning: $$ \tan (\tan^{-1}y - \tan^{-1}x) = \tan K$$ Now let $\tan^{-1}y = \theta,\,\,\,\tan^{-1}x = \phi$, so we have $$\tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi},$$ using addition formulae. Sub in the above conditions and let $\tan K = C$ then rearrange gives the result.