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

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

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(dy)/(dx)=2+sqrt(3+y-2x)

\frac{d y}{d x} = 2 + 3 + y - 2 x

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

Solution steps

\frac{d y}{d x} = 2 + 3 + y - 2 x

Solve using substitution:

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

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

\frac{d}{d x} ((2 x + 5) (3 x - 4))

sin (2 x + 2)

\int co tan (\frac{x}{3}) d x

y^{^{''}} + 9 y = 2 cos (3 t)

x \to 0 lim (- \frac{2}{x})

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