What is the solution for y^{''}-5y^'=28xe^{-2x}+10 ?

The solution for y^{''}-5y^'=28xe^{-2x}+10 is y=c_{1}e^{5x}+c_{2}+2e^{-2x}x+9/7 e^{-2x}-2x

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y^{''}-5y^'=28xe^{-2x}+10

y^{^{''}} - 5 y^{^{'}} = 28 x e^{- 2 x} + 10

y = c_{1} e^{5 x} + c_{2} + 2 e^{- 2 x} x + \frac{9}{7} e^{- 2 x} - 2 x

Solution steps

y^{''} (x) - 5 y^{'} (x) = 28 x e^{- 2 x} + 10

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} + 2 e^{- 2 x} x + \frac{9}{7} e^{- 2 x} - 2 x

y = c_{1} e^{5 x} + c_{2} + 2 e^{- 2 x} x + \frac{9}{7} e^{- 2 x} - 2 x

y^{^{''}} - 9 y^{^{'}} + 20 y = 6 x + e^{4 x}

y^{^{''}} + 9 y = 16 sin (3 x) + 12 cos (3 x)

2 x^{^{''}} + 5 x^{^{'}} + 2 x = 10 sin (t)

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 e^{5} t

y^{^{'''}} - 4 y^{^{''}} - y^{^{'}} + 4 y = e^{6 x}

What is the solution for y^{''}-5y^'=28xe^{-2x}+10 ?
The solution for y^{''}-5y^'=28xe^{-2x}+10 is y=c_{1}e^{5x}+c_{2}+2e^{-2x}x+9/7 e^{-2x}-2x