What is the solution for 2yy^'=12t^2 ?

The solution for 2yy^'=12t^2 is y=sqrt(4t^3+c_{1)},y=-sqrt(4t^3+c_{1)}

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2yy^'=12t^2

2 y y^{^{'}} = 12 t^{2}

y = 4 t^{3} + c_{1}, y = - 4 t^{3} + c_{1}

Solution steps

2 y y^{'} (t) = 12 t^{2}

Solve separable ODE:

y = 4 t^{3} + c_{1}, y = - 4 t^{3} + c_{1}

y = 4 t^{3} + c_{1}, y = - 4 t^{3} + c_{1}

\frac{d y}{d x} = \frac{y ^{3}}{x ^{4}}

\frac{d y}{d x} = - \frac{2}{4} x^{3} y^{8}

\frac{d y}{d t} = \frac{0.002 \cdot ( - 4.43 y )}{10}

x^{3} \frac{d y}{d x} + 4 y^{2} = 0, y (1) = 1

2 x y^{^{'}} ln (x) = - 2 y

What is the solution for 2yy^'=12t^2 ?
The solution for 2yy^'=12t^2 is y=sqrt(4t^3+c_{1)},y=-sqrt(4t^3+c_{1)}