What is the solution for (1-x)(dy)/(dx)-y^2=0 ?

The solution for (1-x)(dy)/(dx)-y^2=0 is y=-1/(-ln(1-x)+c_{1)}

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(1-x)(dy)/(dx)-y^2=0

(1 - x) \frac{d y}{d x} - y^{2} = 0

y = - \frac{1}{- ln ( 1 - x ) + c _{1}}

Solution steps

(1 - x) \frac{d y}{d x} - y^{2} = 0

Solve separable ODE:

y = - \frac{1}{- ln ( 1 - x ) + c _{1}}

y = - \frac{1}{- ln ( 1 - x ) + c _{1}}

\frac{d y}{d x} = x \cdot y

y^{^{'}} = 3 y e^{t}, y (0) = 1

\frac{d y}{d x} = x^{2} y - 1.1 y

y^{^{'}} = \frac{( 3 x ^{2} - 1 )}{( 3 + 2 y )}

\frac{d y}{d t} = \frac{3 t ^{2} + 4 t + 2}{2 ( y - 1 )}, y (0) = - 1

What is the solution for (1-x)(dy)/(dx)-y^2=0 ?
The solution for (1-x)(dy)/(dx)-y^2=0 is y=-1/(-ln(1-x)+c_{1)}