What is the solution for y^'=(x-2)/(y-4) ?

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

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y^'=(x-2)/(y-4)

y^{^{'}} = \frac{x - 2}{y - 4}

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

Solution steps

y^{'} (x) = \frac{x - 2}{y - 4}

Solve separable ODE:

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

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

f (x) = x^{2} + 9, at x = - 4

\int x os (x) d x

n = 1 \sum \infty \frac{e ^{n}}{n}

t y^{^{'}} + 2 y = t^{2} - t + 1, y (1) = 0.5

sin^{2} (14 x)

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