What is the solution for y^'+1/2 (x-2)y^2=0,y(1/10)=6 ?

The solution for y^'+1/2 (x-2)y^2=0,y(1/10)=6 is y=-(1200)/(-300x^2+1200x-317)

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y^'+1/2 (x-2)y^2=0,y(1/10)=6

y^{^{'}} + \frac{1}{2} (x - 2) y^{2} = 0, y (\frac{1}{10}) = 6

y = - \frac{1200}{- 300 x ^{2} + 1200 x - 317}

Solution steps

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

Solve separable ODE:

y = - \frac{1200}{- 300 x ^{2} + 1200 x - 317}

y = - \frac{1200}{- 300 x ^{2} + 1200 x - 317}

(3 y^{2} - 3) d x + (5 x y) d y = 0

f^{^{'}} (x) = 4 f (x)

x - 1 \frac{d y}{d x} = \frac{x}{y}, y = 1, y > 0, x = 2

y^{^{'}} - \frac{y}{1 + e ^{x}} = 0

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

What is the solution for y^'+1/2 (x-2)y^2=0,y(1/10)=6 ?
The solution for y^'+1/2 (x-2)y^2=0,y(1/10)=6 is y=-(1200)/(-300x^2+1200x-317)