Help on this maths.?

Question

Can anyone explain this equation to me?

1 = (y/u) du/dy + u (u / (1+u^2))
1 = (y/u) du/dy + u^2 / (1+u^2)
(y/u) du/dy + u^2 / (1+u^2) - 1 = 0
(y/u) du/dy - 1 / (1+u^2) = 0
y du/dy = u / (1+u^2)

Answer 1

1 = (y/u) du/dy + u (u / (1+u^2))
1 = (y/u) du/dy + u^2 / (1+u^2)
step 1 to 2 is simply multiply u inside and get u^2

1 = (y/u) du/dy + u^2 / (1+u^2)
(y/u) du/dy + u^2 / (1+u^2) - 1 = 0
step 2 to 3 is just simply shift 1 to the right and make the equation to 0

(y/u) du/dy + u^2 / (1+u^2) - 1 = 0
(y/u) du/dy - 1 / (1+u^2) = 0
step 3 to 4: change the -1 to -((1+u^2) / (1+u^2)) and add up to with u^2 / (1+u^2) , therefore u^2 is cancelled and get - 1 / (1+u^2)

(y/u) du/dy - 1 / (1+u^2) = 0
y du/dy = u / (1+u^2)
these step is simply shift the - 1 / (1+u^2) to the right side and then bring the u to the right side

Answer 2

So far it was simple arithmetic and you are good. However, to solve this differential equation you need to continue like this:

dy/y = ((1 + u^2)/u) du = du/u + u du. After integration you get:

ln y = ln u + u^2/2 + C. This can be rewritten as:

y = C*u*e^(u^2/2), where C is a constant.