What is the solution for y^'=(x^2+1)/(2-y),y(-3)=4 ?

The solution for y^'=(x^2+1)/(2-y),y(-3)=4 is y=-(-12-sqrt(-24x^3-72x-720))/6

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y^'=(x^2+1)/(2-y),y(-3)=4

y^{^{'}} = \frac{x ^{2} + 1}{2 - y}, y (- 3) = 4

y = - \frac{- 12 - - 24 x ^{3} - 72 x - 720}{6}

Solution steps

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

Solve separable ODE:

y = - \frac{- 12 - - 24 x ^{3} - 72 x - 720}{6}

y = - \frac{- 12 - - 24 x ^{3} - 72 x - 720}{6}

\frac{d P}{d t} - a P (t) = b

\frac{d y}{d x} = - \frac{2}{4} x^{7} y^{5}

y^{^{'}} + \frac{y}{6} = 2, y (- 1) = 0

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

\frac{d y}{d x} = \frac{e ^{x} - cos ( 2 x )}{y}

What is the solution for y^'=(x^2+1)/(2-y),y(-3)=4 ?
The solution for y^'=(x^2+1)/(2-y),y(-3)=4 is y=-(-12-sqrt(-24x^3-72x-720))/6