What is the solution for y(dy)/(dx)=((9x^3-1))/((7+y^2)) ?

The solution for y(dy)/(dx)=((9x^3-1))/((7+y^2)) is

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y(dy)/(dx)=((9x^3-1))/((7+y^2))

y \frac{d y}{d x} = \frac{( 9 x ^{3} - 1 )}{( 7 + y ^{2} )}

y = - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - 9 x^{4} - 4 x + c_{1} + 49 - 7, y = - - 9 x^{4} - 4 x + c_{1} + 49 - 7

Solution steps

y \frac{d y}{d x} = \frac{( 9 x ^{3} - 1 )}{( 7 + y ^{2} )}

Solve separable ODE:

y = - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - 9 x^{4} - 4 x + c_{1} + 49 - 7, y = - - 9 x^{4} - 4 x + c_{1} + 49 - 7

y = - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - - 7 + 9 x^{4} - 4 x + c_{1} + 49, y = - 9 x^{4} - 4 x + c_{1} + 49 - 7, y = - - 9 x^{4} - 4 x + c_{1} + 49 - 7

\frac{d y}{d x} = (\frac{y ^{2}}{e ^{x}}), y (0) = (\frac{1}{2})

2 cos^{2} (x) d y - y^{3} sin (2 x) d x = 0

2 x y^{2} + x^{2} 2 y \frac{d y}{d x} = 0

y^{^{'}} = e^{x - y}, y (0) = ln (7)

\frac{d y}{d x} + x^{2} y = 5 x^{2}

What is the solution for y(dy)/(dx)=((9x^3-1))/((7+y^2)) ?
The solution for y(dy)/(dx)=((9x^3-1))/((7+y^2)) is