What is the solution for y^'=(6y^2)/(x^2),y(-2)=5 ?

The solution for y^'=(6y^2)/(x^2),y(-2)=5 is y=(5x)/(2(15+8x))

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y^'=(6y^2)/(x^2),y(-2)=5

y^{^{'}} = \frac{6 y ^{2}}{x ^{2}}, y (- 2) = 5

y = \frac{5 x}{2 ( 15 + 8 x )}

Solution steps

y^{'} (x) = \frac{6 y ^{2}}{x ^{2}}

Solve separable ODE:

y = \frac{5 x}{2 ( 15 + 8 x )}

y = \frac{5 x}{2 ( 15 + 8 x )}

\frac{d}{d x} (\frac{4}{3} tan (x))

x \to - 1 lim (13 + x)

\frac{d x}{d t} = \frac{x ^{2} + t t ^{2} + x ^{2}}{t x}

\frac{d}{d x} (\frac{x ^{2}}{4 + x})

\frac{\partial}{\partial x} (- x y - x^{2})

What is the solution for y^'=(6y^2)/(x^2),y(-2)=5 ?
The solution for y^'=(6y^2)/(x^2),y(-2)=5 is y=(5x)/(2(15+8x))