What is the solution for 2yy^'=1,y(0)=sqrt(x-1) ?

The solution for 2yy^'=1,y(0)=sqrt(x-1) is y=sqrt(x+c_{1)},y=-sqrt(x+c_{1)}

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2yy^'=1,y(0)=sqrt(x-1)

2 y y^{^{'}} = 1, y (0) = x - 1

y = x + c_{1}, y = - x + c_{1}

Solution steps

2 y y^{'} (x) = 1

Solve separable ODE:

y = x + c_{1}, y = - x + c_{1}

y = x + c_{1}, y = - x + c_{1}

a (y) \frac{d y}{d t} = - a 2 g y

\frac{d y}{d x} = \frac{3 x ^{2} - 6 x ^{2} y ^{2}}{y - x ^{3} y}

(6 x - 1) y^{^{'}} + 6 y = 0

\frac{d y}{d x} = - \frac{y}{\frac{4320}{7}}

(1 - x^{2}) y d y - x e^{- y} d x = 0, y (0) = 0

What is the solution for 2yy^'=1,y(0)=sqrt(x-1) ?
The solution for 2yy^'=1,y(0)=sqrt(x-1) is y=sqrt(x+c_{1)},y=-sqrt(x+c_{1)}