What is the solution for y^'=(x^2+3)/y ,y(0)=1 ?

The solution for y^'=(x^2+3)/y ,y(0)=1 is y=sqrt((2x^3)/3+6x+1)

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y^'=(x^2+3)/y ,y(0)=1

y^{^{'}} = \frac{x ^{2} + 3}{y}, y (0) = 1

y = \frac{2 x ^{3}}{3} + 6 x + 1

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{3}}{3} + 6 x + 1

y = \frac{2 x ^{3}}{3} + 6 x + 1

x e^{x}

x \to - 4 lim (16 - x^{2})

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

x \to 0 + lim (\frac{2}{x})

x \frac{d y}{d x} + 8 y = e^{x^{8}}

What is the solution for y^'=(x^2+3)/y ,y(0)=1 ?
The solution for y^'=(x^2+3)/y ,y(0)=1 is y=sqrt((2x^3)/3+6x+1)