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

The solution for y^'=(x-1)/(y^2) is y=\sqrt[3]{(3x^2)/2-3x+c_{1}}

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y^'=(x-1)/(y^2)

y^{^{'}} = \frac{x - 1}{y ^{2}}

Solution steps

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

Solve separable ODE:

\frac{d y}{d x} = \frac{( 3 e ^{x} )}{( 3 y + 1 )}

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

\frac{d y}{d x} = 6 - \frac{y}{20}

y^{^{'}} + 3 y = 1, 3 sin (2 t) y (0) = 6

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

What is the solution for y^'=(x-1)/(y^2) ?
The solution for y^'=(x-1)/(y^2) is y=\sqrt[3]{(3x^2)/2-3x+c_{1}}