What is the solution for (y^')/3 =x^2y,y(0)=-8 ?

The solution for (y^')/3 =x^2y,y(0)=-8 is y=-8e^{x^3}

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(y^')/3 =x^2y,y(0)=-8

\frac{y ^{^{'}}}{3} = x^{2} y, y (0) = - 8

y = - 8 e^{x^{3}}

Solution steps

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

Solve linear ODE:

y = - 8 e^{x^{3}}

y = - 8 e^{x^{3}}

x \frac{d x}{d t} + b t = 0

\frac{\partial}{\partial x} (2 y cos^{2} (x))

\int_{0}^{6} 4 x - 2 x^{2} d x

\int x (3 - x^{2})^{2} d x

x \to 3 lim (\frac{1}{x - 3} + 2)

What is the solution for (y^')/3 =x^2y,y(0)=-8 ?
The solution for (y^')/3 =x^2y,y(0)=-8 is y=-8e^{x^3}