What is the solution for (1-u)udw+(1-u)wdu=(1+u)dw ?

The solution for (1-u)udw+(1-u)wdu=(1+u)dw is w=(e^{arctan(u)+c_{1}})/(sqrt(u^2+1))

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(1-u)udw+(1-u)wdu=(1+u)dw

(1 - u) u d w + (1 - u) w d u = (1 + u) d w

w = \frac{e ^{a r c t a n (u) + c_{1}}}{u ^{2} + 1}

Solution steps

(1 - u) u d w + (1 - u) w d u = (1 + u) d w

Solve separable ODE:

w = \frac{e ^{a r c t a n (u) + c_{1}}}{u ^{2} + 1}

w = \frac{e ^{a r c t a n (u) + c_{1}}}{u ^{2} + 1}

\frac{d y}{d x} = \frac{( 9 + 18 x )}{x y ^{2}}, y (1) = 6

\frac{d y}{d x} y = (x + 7) (2 x + 9)

y^{^{'}} (p) = \frac{1}{9} y (p) (2 y (p) - 1) (p - 3)

2 y y^{^{'}} = 15 t^{2}, y (0) = 6

\frac{d y}{d x} + \frac{d y}{d x} + y = 0

What is the solution for (1-u)udw+(1-u)wdu=(1+u)dw ?
The solution for (1-u)udw+(1-u)wdu=(1+u)dw is w=(e^{arctan(u)+c_{1}})/(sqrt(u^2+1))