What is the solution for y^'=-(9x)/y ?

The solution for y^'=-(9x)/y is y=sqrt(-9x^2+c_{1)},y=-sqrt(-9x^2+c_{1)}

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y^'=-(9x)/y

y^{^{'}} = - \frac{9 x}{y}

y = - 9 x^{2} + c_{1}, y = - - 9 x^{2} + c_{1}

Solution steps

y^{'} (x) = - \frac{9 x}{y}

Solve separable ODE:

y = - 9 x^{2} + c_{1}, y = - - 9 x^{2} + c_{1}

y = - 9 x^{2} + c_{1}, y = - - 9 x^{2} + c_{1}

6 (y^{2} + 2 x y \frac{d y}{d x}) + 18 x = - 4 y^{3} \frac{d y}{d x}

y^{^{'}} (x) = \frac{y}{x ( y - 3 )}, y (2) = 6

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

\frac{d x}{d t} + \frac{9}{50} x = 15

y^{5} \frac{d y}{d x} = 5 x^{2} y^{6} - 15 x^{2}

What is the solution for y^'=-(9x)/y ?
The solution for y^'=-(9x)/y is y=sqrt(-9x^2+c_{1)},y=-sqrt(-9x^2+c_{1)}