What is the solution for (dy)/(dx)y=u^2-2u,u= 1/(x^2) ?

The solution for (dy)/(dx)y=u^2-2u,u= 1/(x^2) is y=sqrt(2(u^2-2u)x+c_{1)},y=-sqrt(2(u^2-2u)x+c_{1)}

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(dy)/(dx)y=u^2-2u,u= 1/(x^2)

\frac{d y}{d x} y = u^{2} - 2 u, u = \frac{1}{x ^{2}}

y = 2 (u^{2} - 2 u) x + c_{1}, y = - 2 (u^{2} - 2 u) x + c_{1}

Solution steps

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

Solve separable ODE:

y = 2 (u^{2} - 2 u) x + c_{1}, y = - 2 (u^{2} - 2 u) x + c_{1}

y = 2 (u^{2} - 2 u) x + c_{1}, y = - 2 (u^{2} - 2 u) x + c_{1}

3 = \frac{π}{12} \frac{d}{d t} ((h (t))^{3})

2 sin (3 x) sin (2 y) d y = 3 cos (3 x) cos (2 y) d x

7 x y \frac{d y}{d x} = 3, y > 0, x > 0

\frac{d x}{d t} + \frac{8 x}{240} = 2

\frac{d y}{d x} = 4 x - 7 x y

What is the solution for (dy)/(dx)y=u^2-2u,u= 1/(x^2) ?
The solution for (dy)/(dx)y=u^2-2u,u= 1/(x^2) is y=sqrt(2(u^2-2u)x+c_{1)},y=-sqrt(2(u^2-2u)x+c_{1)}