What is the solution for y^'y=(x^2-2x+2)(4x^3-x^2+5) ?

The solution for y^'y=(x^2-2x+2)(4x^3-x^2+5) is

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y^'y=(x^2-2x+2)(4x^3-x^2+5)

y^{^{'}} y = (x^{2} - 2 x + 2) (4 x^{3} - x^{2} + 5)

y = \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}, y = - \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}

Solution steps

y^{'} (x) y = (x^{2} - 2 x + 2) (4 x^{3} - x^{2} + 5)

Solve separable ODE:

y = \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}, y = - \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}

y = \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}, y = - \frac{4 x ^{6}}{3} - \frac{18 x ^{5}}{5} + 5 x^{4} + 2 x^{3} - 10 x^{2} + 20 x + c_{1}

\frac{d y}{d x} = \frac{cos ( x )}{3 y}

\frac{d y}{d x} y = (x^{6} ln (x))^{4}

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

\frac{d y}{d x} = \frac{x y ^{3} - y ^{3}}{e ^{x}}

y^{^{'}} + \frac{x}{x ^{2} - 9} \cdot y = 0

What is the solution for y^'y=(x^2-2x+2)(4x^3-x^2+5) ?
The solution for y^'y=(x^2-2x+2)(4x^3-x^2+5) is