What is the solution for y^'=((x-3))/((y-4)) ?

The solution for y^'=((x-3))/((y-4)) is y=4+sqrt(x^2-6x+c_{1)+16},y=4-sqrt(x^2-6x+c_{1)+16}

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y^'=((x-3))/((y-4))

y^{^{'}} = \frac{( x - 3 )}{( y - 4 )}

y = 4 + x^{2} - 6 x + c_{1} + 16, y = 4 - x^{2} - 6 x + c_{1} + 16

Solution steps

y^{'} (x) = \frac{( x - 3 )}{( y - 4 )}

Solve separable ODE:

y = 4 + x^{2} - 6 x + c_{1} + 16, y = 4 - x^{2} - 6 x + c_{1} + 16

y = 4 + x^{2} - 6 x + c_{1} + 16, y = 4 - x^{2} - 6 x + c_{1} + 16

\int x^{2} - x - 2 d x

\frac{d}{d x} (\frac{3 x + 1}{5 x ^{2} + 1})

f (x) = \frac{sin ( x )}{x ^{3}}

x y^{^{'}} + y = 6 x^{2}, y (1) = 2

n \to \infty lim (n^{10})

What is the solution for y^'=((x-3))/((y-4)) ?
The solution for y^'=((x-3))/((y-4)) is y=4+sqrt(x^2-6x+c_{1)+16},y=4-sqrt(x^2-6x+c_{1)+16}