What is the solution for y^'=1+sqrt(y-2x+3) ?

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

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y^'=1+sqrt(y-2x+3)

y^{^{'}} = 1 + y - 2 x + 3

y = 2 x + W (e^{\frac{x + c _{1}}{2}})^{2} + 2 W (e^{\frac{x + c _{1}}{2}}) - 2

Solution steps

y^{'} (x) = 1 + y - 2 x + 3

Solve using substitution:

y = 2 x + W (e^{\frac{x + c _{1}}{2}})^{2} + 2 W (e^{\frac{x + c _{1}}{2}}) - 2

y = 2 x + W (e^{\frac{x + c _{1}}{2}})^{2} + 2 W (e^{\frac{x + c _{1}}{2}}) - 2

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