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=sqrt(x^2+2x-1)*integral of (x-1)/((x^2+2x-1)^{3/2)}

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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 = x^{2} + 2 x - 1 \cdot \int \frac{x - 1}{( x ^{2} + 2 x - 1 ) ^{\frac{3}{2}}} d x

Solution steps

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

Solve linear ODE:

y = x^{2} + 2 x - 1 \cdot \int \frac{x - 1}{( x ^{2} + 2 x - 1 ) ^{\frac{3}{2}}} d x

y = x^{2} + 2 x - 1 \cdot \int \frac{x - 1}{( x ^{2} + 2 x - 1 ) ^{\frac{3}{2}}} d x

x \to 0 lim (x^{4} cos (\frac{5}{x}))

\int t^{2} (t^{3} - 4)^{2} d t

(x - 2)^{4} + e^{x - 2} + (x - 2 \cdot z)^{2}

\frac{d y}{d x} = x + 7 y

\int 27 x^{26} d x

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=sqrt(x^2+2x-1)*integral of (x-1)/((x^2+2x-1)^{3/2)}