What is the solution for 3y^2(x-3)(dy)/(dx)=k ?

The solution for 3y^2(x-3)(dy)/(dx)=k is y=\sqrt[3]{kln(x-3)+c_{1}}

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3y^2(x-3)(dy)/(dx)=k

3 y^{2} (x - 3) \frac{d y}{d x} = k

y = 3 k ln (x - 3) + c_{1}

Solution steps

3 y^{2} (x - 3) \frac{d y}{d x} = k

Solve separable ODE:

y = 3 k ln (x - 3) + c_{1}

y = 3 k ln (x - 3) + c_{1}

\frac{d x}{d y} = tan (x)

y^{^{'}} = \frac{e ^{- x}}{sin ( y )}, y (0) = 1

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

y^{^{'}} = x^{2} y^{- 3}, y (0) = 4

\frac{d y}{d x} y = x^{6} sin (x)

What is the solution for 3y^2(x-3)(dy)/(dx)=k ?
The solution for 3y^2(x-3)(dy)/(dx)=k is y=\sqrt[3]{kln(x-3)+c_{1}}