What is the solution for y^'=x^2y^{-7} ?

The solution for y^'=x^2y^{-7} is y=\sqrt[8]{(8x^3)/3+c_{1}},y=-\sqrt[8]{(8x^3)/3+c_{1}}

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y^'=x^2y^{-7}

y^{^{'}} = x^{2} y^{- 7}

y = 8 \frac{8 x ^{3}}{3} + c_{1}, y = - 8 \frac{8 x ^{3}}{3} + c_{1}

Solution steps

y^{'} (x) = x^{2} y^{- 7}

Solve separable ODE:

y = 8 \frac{8 x ^{3}}{3} + c_{1}, y = - 8 \frac{8 x ^{3}}{3} + c_{1}

y = 8 \frac{8 x ^{3}}{3} + c_{1}, y = - 8 \frac{8 x ^{3}}{3} + c_{1}

x^{3} d y + x^{2} y d x + y d x = x d y

y^{^{'}} + 4 y = 0, y (0) = 5

y^{^{'}} - x^{2} y = y e^{- x}

\frac{d y}{d x} = \frac{1}{8} (y^{\frac{1}{2}}) (cos (y^{\frac{1}{2}}))^{2}

\frac{x ^{2} d y - y ^{2} d x}{( x - y ) ^{2}} = 0

What is the solution for y^'=x^2y^{-7} ?
The solution for y^'=x^2y^{-7} is y=\sqrt[8]{(8x^3)/3+c_{1}},y=-\sqrt[8]{(8x^3)/3+c_{1}}