How can I solve this tricky integral in my differential equation?

[VanHa]

I've been trying to solve differential equation:

[tex] y' = \frac{x+y}{x-y} [/tex]

I came to the point where I got following integrals:

[tex]\int \frac{(1-z) \cdot \,dz}{1+z^2} = \int \frac{dx}{x} [/tex]

The integral on the left side is the problem. I tried substitution:
[tex]t = 1+z^2 [/tex]
but I always end up with one dz left in the numerator.

I did the differential equation with numerator and denumerator inversed without problems, but I'm stuck on this one and I have a feeling that I can't figure out a trivial thing.

Any hints?

Thanks for help!

 

[marksman95]

Try substitution y = t·x right at the beginning. (it's an homogeneous differential equation)

EDIT: Sorry for my stupidity. You already did. I'm trying to integrate your expression right now.

 

[marksman95]

OK. Try separating the fraction into two fractions and integrating them separetly (you have two summands at the numerator, so you can express that as the sum of two separated fraction with the same denominator).

 

[throneoo]

seems that a trig substitution would work.. have you tried z=tan(u)?

 

[SteamKing]

What's the definition of z in the OP?

In any event, splitting the z-integral into two parts makes it rather easy to integrate, as suggested by marksman95. You should be able to find one integral in a table of integrals and the other integral can be solved by substitution.

 

[VanHa]

Thanks for help people. I will try both the trig substitution and by splitting the integral, and see what I will get. I will notice you on how it went. Thanks again.