What is the solution for y^{''}+2y^'=24x+e^{-2x} ?

The solution for y^{''}+2y^'=24x+e^{-2x} is y=c_{1}+c_{2}e^{-2x}+6x^2-6x-(e^{-2x}x)/2

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y^{''}+2y^'=24x+e^{-2x}

y^{^{''}} + 2 y^{^{'}} = 24 x + e^{- 2 x}

y = c_{1} + c_{2} e^{- 2 x} + 6 x^{2} - 6 x - \frac{e ^{- 2 x} x}{2}

Solution steps

y^{''} (x) + 2 y^{'} (x) = 24 x + e^{- 2 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + 6 x^{2} - 6 x - \frac{e ^{- 2 x} x}{2}

y = c_{1} + c_{2} e^{- 2 x} + 6 x^{2} - 6 x - \frac{e ^{- 2 x} x}{2}

x^{^{''}} + 16 x = sin (t)

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} - 20 y = 5 e^{- 3 t}

y^{^{''}} + y^{^{'}} + 4 = - 20 x

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 3

y^{^{''}} - 6 y^{^{'}} + 9 y = 15 x + 2

What is the solution for y^{''}+2y^'=24x+e^{-2x} ?
The solution for y^{''}+2y^'=24x+e^{-2x} is y=c_{1}+c_{2}e^{-2x}+6x^2-6x-(e^{-2x}x)/2