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

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

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(x^2+2x-1)y^'-(x+1)y=x+1

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

y = c_{1} x^{2} + 2 x - 1 - 1

Solution steps

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

Solve separable ODE:

y = c_{1} x^{2} + 2 x - 1 - 1

y = c_{1} x^{2} + 2 x - 1 - 1

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

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

y^{^{'}} - 2 x y^{3} = 0, y (2) = 1

\frac{d y}{d t} = 0.029 y (1 - \frac{y}{5})

\frac{d x}{d t} = 3 - \frac{x}{100}

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