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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + x^{^{'}} + x = 1, x^{^{'}} (0) = 0

y^{^{''}} - 8 y^{^{'}} + 4 y = 8 e^{2 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 16 e^{3 x}

y^{^{''}} + 2 y^{^{'}} + y = t, y (0) = t, y^{^{'}} (0) = 2

y^{^{''}} + y = 2 cos (t)

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