What is the solution for (dy}{dx}=\frac{x^3+1)/y ,y=2,x=1 ?

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

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(dy}{dx}=\frac{x^3+1)/y ,y=2,x=1

\frac{d y}{d x} = \frac{x ^{3} + 1}{y}, y = 2, x = 1

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

Solution steps

\frac{d y}{d x} = \frac{x ^{3} + 1}{y}

Solve separable ODE:

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

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

\frac{d y}{d x} y - 6 x^{3} + cos (x) = 0

y^{^{'}} = 3 y - 15

y^{^{'}} = \frac{x y ^{3}}{5}, y (0) = 5

\frac{d x}{d y} = \frac{4}{x e ^{y}}

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

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