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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y = - 48 t^{2} e^{3 t}

d, d (t) = \frac{100 ln ( t + 2 )}{t + 2}

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 e^{- x} - 10 cos (3 x)

y^{^{''}} - 81 y = sinh (9 x)

y^{^{''}} + 3 y^{^{'}} = 1

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