What is the solution for y^{''}+3y^'-4y=5e^{-4x} ?

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

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y^{''}+3y^'-4y=5e^{-4x}

y^{^{''}} + 3 y^{^{'}} - 4 y = 5 e^{- 4 x}

y = c_{1} e^{x} + c_{2} e^{- 4 x} - e^{- 4 x} x

Solution steps

y^{''} (x) + 3 y^{'} (x) - 4 y = 5 e^{- 4 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} - e^{- 4 x} x

y = c_{1} e^{x} + c_{2} e^{- 4 x} - e^{- 4 x} x

y^{^{''}} + 5 y^{^{'}} + 17 y = 11 cos (3 x)

y^{^{''}} - 8 y^{^{'}} - 15 y = - 6 cos (5 x)

x^{^{''''}} + x^{^{'''}} - 4 x^{^{''}} + 6 x^{^{'}} = 12

y^{^{''}} - 6 y^{^{'}} + 8 y = 85 sin (x)

x^{^{''}} + 9 x = 2 (t - 1) \cdot 2 (t - 2), x (0) = 0, x^{^{'}} (0) = 0

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