What is the solution for y^3y^'+x^3=0 ?

The solution for y^3y^'+x^3=0 is y=\sqrt[4]{-x^4+c_{1}},y=-\sqrt[4]{-x^4+c_{1}}

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y^3y^'+x^3=0

y^{3} y^{^{'}} + x^{3} = 0

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

Solution steps

y^{3} y^{'} (x) + x^{3} = 0

Solve separable ODE:

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

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

x^{2} + 2 x

\int (6 x^{2} - 8 x + 2) d x

y^{^{'''}} + y^{^{'}} = cot (x)

y^{^{''}} + 5 \frac{d y}{d x} + 6 y = 2 e^{x}

n = 1 \sum \infty \frac{9 n}{2 n + 1}

What is the solution for y^3y^'+x^3=0 ?
The solution for y^3y^'+x^3=0 is y=\sqrt[4]{-x^4+c_{1}},y=-\sqrt[4]{-x^4+c_{1}}