What is the solution for 2yy^'=1+(y^')2 ?

The solution for 2yy^'=1+(y^')2 is y=1+sqrt(t+c_{1)+1},y=1-sqrt(t+c_{1)+1}

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2yy^'=1+(y^')2

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

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

Solution steps

2 y y^{'} (t) = 1 + (y^{'} (t)) 2

Solve separable ODE:

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

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

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

\frac{x ^{3} + 1}{x ^{2}} \frac{d y}{d x} = \frac{1}{y}

y y^{^{'}} = - 5 x

\frac{d y}{d x} - \frac{2 x}{x ^{2} + 1} y = - \frac{2 x}{x ^{2} + 1}

(- s^{2} - s) y^{^{'}} (s) + (- 3 s - 1) y (s) = 0

What is the solution for 2yy^'=1+(y^')2 ?
The solution for 2yy^'=1+(y^')2 is y=1+sqrt(t+c_{1)+1},y=1-sqrt(t+c_{1)+1}