What is the solution for y^'=2+sqrt(y-2x+3) ?

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

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y^'=2+sqrt(y-2x+3)

y^{^{'}} = 2 + y - 2 x + 3

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

Solution steps

y^{'} (x) = 2 + y - 2 x + 3

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} (\frac{2}{( x + 3 ) ^{3}})

y

2

- csc (x) cot (x)

10 x + \frac{d y}{d x} = 0

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