What is the solution for (x^2-1)y^'+xy(1-y)=0 ?

The solution for (x^2-1)y^'+xy(1-y)=0 is y=-(e^{c_{1}}((x^2-1)^{1/2}+e^{c_{1}}))/(x^2-1-e^{2c_{1)}}

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(x^2-1)y^'+xy(1-y)=0

(x^{2} - 1) y^{^{'}} + x y (1 - y) = 0

y = - \frac{e ^{c_{1}} ( ( x ^{2} - 1 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} - 1 - e ^{2 c_{1}}}

Solution steps

(x^{2} - 1) y^{'} (x) + x y (1 - y) = 0

Solve separable ODE:

y = - \frac{e ^{c_{1}} ( ( x ^{2} - 1 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} - 1 - e ^{2 c_{1}}}

y = - \frac{e ^{c_{1}} ( ( x ^{2} - 1 ) ^{\frac{1}{2}} + e ^{c_{1}} )}{x ^{2} - 1 - e ^{2 c_{1}}}

\frac{d x}{d t} = e^{x} t cos (t)

\frac{d y}{d x} + 9 x^{2} y = 27 x^{2}

\frac{d x}{d t} = 18 - \frac{18}{100} x

y^{^{'}} = - \frac{x - 2}{y - 6}, y (3) = 0

3.03 y^{^{'}} + 0.755 y = 6.05

What is the solution for (x^2-1)y^'+xy(1-y)=0 ?
The solution for (x^2-1)y^'+xy(1-y)=0 is y=-(e^{c_{1}}((x^2-1)^{1/2}+e^{c_{1}}))/(x^2-1-e^{2c_{1)}}