What is the solution for y^'y^2=(t-2)^2 ?

The solution for y^'y^2=(t-2)^2 is y=\sqrt[3]{t^3-6t^2+12t+c_{1}}

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y^'y^2=(t-2)^2

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

Solution steps

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

Solve separable ODE:

y^{^{'}} - \frac{1}{3} a^{2} y = 0

\frac{d y}{d x} = 10 y - x y + 3 y (x^{2} + 49)

(2 x y^{2} + y^{2} + 3 + 6 x) - (4 x^{2} y + 4 x y) y^{^{'}} = 0

\frac{d y}{d x} = 7 e^{y}

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

What is the solution for y^'y^2=(t-2)^2 ?
The solution for y^'y^2=(t-2)^2 is y=\sqrt[3]{t^3-6t^2+12t+c_{1}}