What is the solution for (dy)/(dx)=sqrt(3x+y)-3 ?

The solution for (dy)/(dx)=sqrt(3x+y)-3 is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-3x

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(dy)/(dx)=sqrt(3x+y)-3

\frac{d y}{d x} = 3 x + y - 3

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 3 x

Solution steps

\frac{d y}{d x} = 3 x + y - 3

Solve using substitution:

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 3 x

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 3 x

\int_{0}^{10} x e^{- 0.6 x} d x

\frac{\partial}{\partial y} (y (2 - x - y))

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x y^{^{'}} - y^{^{'}} - (x - 1) sin (x) + y = 0

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What is the solution for (dy)/(dx)=sqrt(3x+y)-3 ?
The solution for (dy)/(dx)=sqrt(3x+y)-3 is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-3x