What is the solution for 2y(y^')+1=e^{x-1} ?

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

(2 x - 1) = - (3 y + 7) y^{^{'}}

y^{^{'}} = 6 y + 1

\frac{d z}{d t} = 4 t e^{3 z}

- 2 x e^{\frac{- H ( x ) x}{x}} d x + x^{2} d H (x) = 0

\frac{d y}{d t} y = arccot (4 t)

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