What is the solution for y^'=((y+1))/((x-1)) ?

The solution for y^'=((y+1))/((x-1)) is y=e^{c_{1}}x-e^{c_{1}}-1

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y^'=((y+1))/((x-1))

y^{^{'}} = \frac{( y + 1 )}{( x - 1 )}

y = e^{c_{1}} x - e^{c_{1}} - 1

Solution steps

y^{'} (x) = \frac{( y + 1 )}{( x - 1 )}

Solve separable ODE:

y = e^{c_{1}} x - e^{c_{1}} - 1

y = e^{c_{1}} x - e^{c_{1}} - 1

y^{^{'}} = x (y^{2} + 1)

2 x y + 6 x + (x^{2} - 4) y^{^{'}} = 0, y (3) = 2

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}, y (0) = \frac{π}{4}

\frac{d y}{d x} = x^{5} e^{- y}

2 (x + 1) y y^{^{'}} - y^{2} = 2, y (5) = 2

What is the solution for y^'=((y+1))/((x-1)) ?
The solution for y^'=((y+1))/((x-1)) is y=e^{c_{1}}x-e^{c_{1}}-1