What is the solution for y^'=(x-1)/(y-4) ?

The solution for y^'=(x-1)/(y-4) is y=4+sqrt(x^2-2x+c_{1)+16},y=4-sqrt(x^2-2x+c_{1)+16}

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y^'=(x-1)/(y-4)

y^{^{'}} = \frac{x - 1}{y - 4}

y = 4 + x^{2} - 2 x + c_{1} + 16, y = 4 - x^{2} - 2 x + c_{1} + 16

Solution steps

y^{'} (x) = \frac{x - 1}{y - 4}

Solve separable ODE:

y = 4 + x^{2} - 2 x + c_{1} + 16, y = 4 - x^{2} - 2 x + c_{1} + 16

y = 4 + x^{2} - 2 x + c_{1} + 16, y = 4 - x^{2} - 2 x + c_{1} + 16

\frac{\partial}{\partial y} (3 y^{2} + 3 x^{2} - 12 y)

y^{^{'}} = 3 e^{- 2 x}

f (x) = \frac{1}{x}, (1, 3)

\frac{\partial}{\partial x} (sin (- 3 x + 1 y + z))

\int (\frac{2 x ^{5} - 4 x + 8 x}{x}) d x

What is the solution for y^'=(x-1)/(y-4) ?
The solution for y^'=(x-1)/(y-4) is y=4+sqrt(x^2-2x+c_{1)+16},y=4-sqrt(x^2-2x+c_{1)+16}