What is the solution for (dy)/(dx)=1+sqrt(y-x) ?

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

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(dy)/(dx)=1+sqrt(y-x)

\frac{d y}{d x} = 1 + y - x

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

Solution steps

\frac{d y}{d x} = 1 + y - x

Solve using substitution:

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

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

3 x^{2} - 6 x - 9, at x = 3

\frac{\partial}{\partial y} (\frac{x y ^{2}}{x ^{2} y ^{3} + 1})

x \to 1 lim (\frac{x + 3}{x - 2})

6, 0, 16 - 2 x

\frac{d y}{d x} = \frac{x y}{3}

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