Simplifying dy/(y+z) = dz/(y-z) for integration

Question

This is a part of a textbook solution. I haven't understood how they simplified dy/(y+z) = dz/(y-z) and got their answer.

I just did (y-z)dy = (y+z)dz and integrated it which gave me z^2 - y^2 + 4yz = c1

Can anyone explain please?

Answer 1

Given that $$ {dy \over y + z} = {dz \over y - z} $$ which yields $$ (y - z) dy = (y + z) dz $$

Simplifying, we get $$ y dy - z dy = y dz + z dz $$ or $$ y dy - z dz = y dz + z dy = d(yz) $$

Rearranging terms, we get $$ y dy - z dz - d(yz) = 0 $$

Integrating both sides, we get $$ {y^2 \over 2} - {z^2 \over 2} - y z = k $$ where $k$ is an integration constant.

Thus, $$ y^2 - z^2 - 2 y z = C_1, $$ where $C_1 = 2 k$ is an integration constant.