What is the solution for y^'=(x-4)/(y-6) ?

The solution for y^'=(x-4)/(y-6) is y=6+sqrt(x^2-8x+c_{1)+36},y=6-sqrt(x^2-8x+c_{1)+36}

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y^'=(x-4)/(y-6)

y^{^{'}} = \frac{x - 4}{y - 6}

y = 6 + x^{2} - 8 x + c_{1} + 36, y = 6 - x^{2} - 8 x + c_{1} + 36

Solution steps

y^{'} (x) = \frac{x - 4}{y - 6}

Solve separable ODE:

y = 6 + x^{2} - 8 x + c_{1} + 36, y = 6 - x^{2} - 8 x + c_{1} + 36

y = 6 + x^{2} - 8 x + c_{1} + 36, y = 6 - x^{2} - 8 x + c_{1} + 36

(4, 5), (1, 2)

x \to 0 lim (\frac{∣ 2 x ∣}{x})

\frac{\partial}{\partial x} (\frac{1}{t ^{2} + 1})

y = \frac{2}{x ^{2}}, (2, \frac{2}{49})

x = y^{2} - 6 y, x = 3 y - y^{2}

What is the solution for y^'=(x-4)/(y-6) ?
The solution for y^'=(x-4)/(y-6) is y=6+sqrt(x^2-8x+c_{1)+36},y=6-sqrt(x^2-8x+c_{1)+36}