What is the solution for (dy)/(dx)=(x^2+3)/(3y) ?

The solution for (dy)/(dx)=(x^2+3)/(3y) is y=(sqrt(18x+2x^3+c_{1)})/3 ,y=-(sqrt(18x+2x^3+c_{1)})/3

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(dy)/(dx)=(x^2+3)/(3y)

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

y = \frac{18 x + 2 x ^{3} + c _{1}}{3}, y = - \frac{18 x + 2 x ^{3} + c _{1}}{3}

Solution steps

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

Solve separable ODE:

y = \frac{18 x + 2 x ^{3} + c _{1}}{3}, y = - \frac{18 x + 2 x ^{3} + c _{1}}{3}

y = \frac{18 x + 2 x ^{3} + c _{1}}{3}, y = - \frac{18 x + 2 x ^{3} + c _{1}}{3}

d er i v a t i v e f (x) = \frac{ln ( x )}{e ^{x}}

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

\frac{\partial}{\partial x} (x cos (4 x y))

\int 13 d x

t a y l or \frac{1}{2 + 7 x}

What is the solution for (dy)/(dx)=(x^2+3)/(3y) ?
The solution for (dy)/(dx)=(x^2+3)/(3y) is y=(sqrt(18x+2x^3+c_{1)})/3 ,y=-(sqrt(18x+2x^3+c_{1)})/3