What is the solution for y^{''}+y^'-2y=3xe^{4x} ?

The solution for y^{''}+y^'-2y=3xe^{4x} is y=c_{1}e^x+c_{2}e^{-2x}+(e^{4x}x)/6-1/12 e^{4x}

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y^{''}+y^'-2y=3xe^{4x}

y^{^{''}} + y^{^{'}} - 2 y = 3 x e^{4 x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{4 x} x}{6} - \frac{1}{12} e^{4 x}

Solution steps

y^{''} (x) + y^{'} (x) - 2 y = 3 x e^{4 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{4 x} x}{6} - \frac{1}{12} e^{4 x}

y = c_{1} e^{x} + c_{2} e^{- 2 x} + \frac{e ^{4 x} x}{6} - \frac{1}{12} e^{4 x}

y^{^{''}} - 16 y = 2 e^{4 t}

y^{^{''}} + 4 y = 3 csc (t)

x^{^{''}} + 2 x^{^{'}} + x = e^{- 2 t}, x (0) = 2, x^{^{'}} (0) = 1

(D - 2)^{2} y = 20 - 3 x e^{2 x}

y^{^{''}} - y = cos^{2} (x)

What is the solution for y^{''}+y^'-2y=3xe^{4x} ?
The solution for y^{''}+y^'-2y=3xe^{4x} is y=c_{1}e^x+c_{2}e^{-2x}+(e^{4x}x)/6-1/12 e^{4x}