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

The solution for (dy}{dx}=\frac{x^3+1)/y is y=sqrt((x^4)/2+2x+c_{1)},y=-sqrt((x^4)/2+2x+c_{1)}

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

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

y = \frac{x ^{4}}{2} + 2 x + c_{1}, y = - \frac{x ^{4}}{2} + 2 x + c_{1}

Solution steps

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

Solve separable ODE:

y = \frac{x ^{4}}{2} + 2 x + c_{1}, y = - \frac{x ^{4}}{2} + 2 x + c_{1}

y = \frac{x ^{4}}{2} + 2 x + c_{1}, y = - \frac{x ^{4}}{2} + 2 x + c_{1}

x^{2} y^{^{'}} = y^{2}

y^{^{'}} = \frac{cos ^{2} ( x )}{y}

x^{2} y^{^{'}} + \frac{1}{y ^{2}} = 0

y^{2} (x - 3) d x - x^{3} d y = 0

\frac{d y}{d t} = \frac{y + 3}{1 - y}

What is the solution for (dy}{dx}=\frac{x^3+1)/y ?
The solution for (dy}{dx}=\frac{x^3+1)/y is y=sqrt((x^4)/2+2x+c_{1)},y=-sqrt((x^4)/2+2x+c_{1)}