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

\int - \frac{x}{x ^{2} + 1} d x

\int \frac{1}{( x - 2 ) ^{2} ( x ^{2} - 4 x + 3 )} d x

f (x) = x^{3} + 4 x^{2} + 3 x

\int \frac{9 - x ^{2}}{x} d x

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

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