What is the solution for (u-1)dx+(x+ux)du=0 ?

The solution for (u-1)dx+(x+ux)du=0 is x=(e^{-u+c_{1}})/(u^2-2u+1)

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(u-1)dx+(x+ux)du=0

(u - 1) d x + (x + ux) d u = 0

x = \frac{e ^{- u + c_{1}}}{u ^{2} - 2 u + 1}

Solution steps

(u - 1) d x + (x + ux) d u = 0

Solve separable ODE:

x = \frac{e ^{- u + c_{1}}}{u ^{2} - 2 u + 1}

x = \frac{e ^{- u + c_{1}}}{u ^{2} - 2 u + 1}

2 t (\frac{d y}{d t}) = 6 y

\frac{d}{d x} (y^{2}) = x^{2}

\frac{d y}{d x} 2 x y = 1

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

\frac{d y}{d x} = 8 x^{6} y - y, (1) = - 2

What is the solution for (u-1)dx+(x+ux)du=0 ?
The solution for (u-1)dx+(x+ux)du=0 is x=(e^{-u+c_{1}})/(u^2-2u+1)