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

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)}