What is the solution for (x^2+7)y^'=x(y^2+9) ?

The solution for (x^2+7)y^'=x(y^2+9) is y=3tan(3/2 ln(x^2+7)+c_{1})

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(x^2+7)y^'=x(y^2+9)

(x^{2} + 7) y^{^{'}} = x (y^{2} + 9)

y = 3 tan (\frac{3}{2} ln (x^{2} + 7) + c_{1})

Solution steps

(x^{2} + 7) y^{'} (x) = x (y^{2} + 9)

Solve separable ODE:

y = 3 tan (\frac{3}{2} ln (x^{2} + 7) + c_{1})

y = 3 tan (\frac{3}{2} ln (x^{2} + 7) + c_{1})

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

\frac{d y}{d x} = 55 y x^{10}

\frac{d v}{d t} = 9 v, v (0) = 11.3

y^{^{'}} = cot (y) sin (x)

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

What is the solution for (x^2+7)y^'=x(y^2+9) ?
The solution for (x^2+7)y^'=x(y^2+9) is y=3tan(3/2 ln(x^2+7)+c_{1})