What is the solution for y^'=((x-2)^2)/y ,y(0)=1 ?

The solution for y^'=((x-2)^2)/y ,y(0)=1 is y=sqrt((2x^3)/3-4x^2+8x+1)

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y^'=((x-2)^2)/y ,y(0)=1

y^{^{'}} = \frac{( x - 2 ) ^{2}}{y}, y (0) = 1

y = \frac{2 x ^{3}}{3} - 4 x^{2} + 8 x + 1

Solution steps

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

Solve separable ODE:

y = \frac{2 x ^{3}}{3} - 4 x^{2} + 8 x + 1

y = \frac{2 x ^{3}}{3} - 4 x^{2} + 8 x + 1

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

\frac{d y}{d x} - \frac{y}{x} = 0

d y = (e^{3 x + 2 y}) d x

x \frac{d y}{d x} = e^{10 y + x}

\frac{d y}{d x} = (x y)

What is the solution for y^'=((x-2)^2)/y ,y(0)=1 ?
The solution for y^'=((x-2)^2)/y ,y(0)=1 is y=sqrt((2x^3)/3-4x^2+8x+1)