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

The solution for (x+1)y^'+(2x-1)y=e^{-2x} is y=-1/(3e^{2x)}+(c_{1}(x^3+3x^2+3x+1))/(e^{2x+2+c_{1)}}

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

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

y = - \frac{1}{3 e ^{2 x}} + \frac{c _{1} ( x ^{3} + 3 x ^{2} + 3 x + 1 )}{e ^{2 x + 2 + c_{1}}}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{3 e ^{2 x}} + \frac{c _{1} ( x ^{3} + 3 x ^{2} + 3 x + 1 )}{e ^{2 x + 2 + c_{1}}}

y = - \frac{1}{3 e ^{2 x}} + \frac{c _{1} ( x ^{3} + 3 x ^{2} + 3 x + 1 )}{e ^{2 x + 2 + c_{1}}}

\frac{d}{d x} (- 6 sin (6 x))

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

arcsin (\frac{x}{7}), at x = 4

\frac{d}{d x} (ln (8 - 9 x))

y^{^{''}} + 4 y = t sin (2 t)

What is the solution for (x+1)y^'+(2x-1)y=e^{-2x} ?
The solution for (x+1)y^'+(2x-1)y=e^{-2x} is y=-1/(3e^{2x)}+(c_{1}(x^3+3x^2+3x+1))/(e^{2x+2+c_{1)}}