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

The solution for y^'=sqrt(9x+y)-9 is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-9x

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y^'=sqrt(9x+y)-9

y^{^{'}} = 9 x + y - 9

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 9 x

Solution steps

y^{'} (x) = 9 x + y - 9

Solve using substitution:

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 9 x

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - 9 x

\int \frac{x}{x - 6} d x

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

\frac{d}{d x} (x^{3} e^{4 x})

t^{2} y^{^{''}} + 7 t y^{^{'}} + 5 y = t

\frac{d}{d x} (\frac{2}{5} x^{15})

What is the solution for y^'=sqrt(9x+y)-9 ?
The solution for y^'=sqrt(9x+y)-9 is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-9x