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

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

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

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

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

Solution steps

\frac{d y}{d x} = 2 - 2 x - y + 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{\partial}{\partial x} (x^{3} \cdot y^{2})

\int_{1}^{0} (- 9 x) d x

\frac{\partial}{\partial t} (ln (x + 2 c t))

\int_{- 1}^{2} ∣ 2 x - 3 ∣ d x

y^{^{''}} + 36 y = 0

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