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

The solution for y^{''}-y^'-2y=3e^{2x}-4e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-x}+e^{2x}x-e^{3x}

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x - e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x - e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + e^{2 x} x - e^{3 x}

x^{^{''}} + 3 x^{^{'}} + 33 x = 11 cos (5 t)

y^{^{''}} + 10 y^{^{'}} + 16 y = 12 cos (2 x) + 4 sin (2 x)

y^{^{'''}} - y^{^{''}} - 12 y^{^{'}} = x - 2 x e^{- 3 x}

y^{^{''}} + y = u_{3} (t), y (0) = 0, y^{^{'}} (0) = 1

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

What is the solution for y^{''}-y^'-2y=3e^{2x}-4e^{3x} ?
The solution for y^{''}-y^'-2y=3e^{2x}-4e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-x}+e^{2x}x-e^{3x}