What is the solution for y^{''}+y^'-12y=xe^{2x} ?

The solution for y^{''}+y^'-12y=xe^{2x} is y=c_{1}e^{3x}+c_{2}e^{-4x}-(e^{2x}x)/6-5/36 e^{2x}

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y^{''}+y^'-12y=xe^{2x}

y^{^{''}} + y^{^{'}} - 12 y = x e^{2 x}

y = c_{1} e^{3 x} + c_{2} e^{- 4 x} - \frac{e ^{2 x} x}{6} - \frac{5}{36} e^{2 x}

Solution steps

y^{''} (x) + y^{'} (x) - 12 y = x e^{2 x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 4 x} - \frac{e ^{2 x} x}{6} - \frac{5}{36} e^{2 x}

y = c_{1} e^{3 x} + c_{2} e^{- 4 x} - \frac{e ^{2 x} x}{6} - \frac{5}{36} e^{2 x}

x^{^{''}} - 2 x^{^{'}} - x = e^{2 t}

y^{^{''}} + 25 y = e^{- t}

y^{^{''}} - y = e^{x} cos (x), y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 6} e^{5 t}

What is the solution for y^{''}+y^'-12y=xe^{2x} ?
The solution for y^{''}+y^'-12y=xe^{2x} is y=c_{1}e^{3x}+c_{2}e^{-4x}-(e^{2x}x)/6-5/36 e^{2x}