What is the solution for y^{''}-4y^'=6e^{3x}-3e^{-x} ?

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

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y^{''}-4y^'=6e^{3x}-3e^{-x}

y^{^{''}} - 4 y^{^{'}} = 6 e^{3 x} - 3 e^{- x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) = 6 e^{3 x} - 3 e^{- x}

Solve linear ODE:

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

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

2 y^{^{''}} + 8 y^{^{'}} - y = 16

y^{^{'''}} + 4 y^{^{'}} = tan (2 x)

y^{^{''}} + y^{^{'}} = e^{5 x}

y^{^{''}} - 16 y^{^{'}} + 64 y = 6 e^{8 x}

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

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