What is the solution for y^'=-(x-2)/(y-6),y(3)=0 ?

The solution for y^'=-(x-2)/(y-6),y(3)=0 is y=6-sqrt(-x^2+4x+33)

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y^'=-(x-2)/(y-6),y(3)=0

y^{^{'}} = - \frac{x - 2}{y - 6}, y (3) = 0

y = 6 - - x^{2} + 4 x + 33

Solution steps

y^{'} (x) = - \frac{x - 2}{y - 6}

Solve separable ODE:

y = 6 - - x^{2} + 4 x + 33

y = 6 - - x^{2} + 4 x + 33

3.03 y^{^{'}} + 0.755 y = 6.05

3 (1 + x) \frac{d y}{d x} = 2 x y (y^{3} - 1)

\frac{d x}{3 sin ( y )} = \frac{4 d y}{5 sin ( x )}

x e^{- y} d y + \frac{x ^{2} + 1}{y} d x = 0

y^{^{'}} y^{1} = - 2 e^{x} y^{2}

What is the solution for y^'=-(x-2)/(y-6),y(3)=0 ?
The solution for y^'=-(x-2)/(y-6),y(3)=0 is y=6-sqrt(-x^2+4x+33)