What is the solution for y^'=(x-1)/(y-3) ?

The solution for y^'=(x-1)/(y-3) is y=3+sqrt(x^2-2x+c_{1)+9},y=3-sqrt(x^2-2x+c_{1)+9}

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y^'=(x-1)/(y-3)

y^{^{'}} = \frac{x - 1}{y - 3}

y = 3 + x^{2} - 2 x + c_{1} + 9, y = 3 - x^{2} - 2 x + c_{1} + 9

Solution steps

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

Solve separable ODE:

y = 3 + x^{2} - 2 x + c_{1} + 9, y = 3 - x^{2} - 2 x + c_{1} + 9

y = 3 + x^{2} - 2 x + c_{1} + 9, y = 3 - x^{2} - 2 x + c_{1} + 9

\int \frac{x ln ( 1 + x ^{2} )}{1 + x ^{2}} d x

\int_{0}^{l n (2)} x e^{x} d x

\frac{\partial}{\partial x} (z \cdot cos (y \cdot x))

y^{^{'}} = 2 y + e^{x}

4 cos (cos (x))

What is the solution for y^'=(x-1)/(y-3) ?
The solution for y^'=(x-1)/(y-3) is y=3+sqrt(x^2-2x+c_{1)+9},y=3-sqrt(x^2-2x+c_{1)+9}