What is the solution for (dy)/(dx)=((x-3))/(y+1) ?

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

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(dy)/(dx)=((x-3))/(y+1)

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{\partial}{\partial y} (x e^{\frac{y}{z}})

\frac{d}{d x} (\frac{3}{2 - x})

\int sec^{2} (3 x + 5) d x

x \to 0 lim (2 + 3 e^{- \frac{3}{x}})

\frac{d}{d x} (2 sin (\frac{1}{2} x))

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