What is the solution for y^{'''}-4y^{''}+y^'+6y=3e^{4t} ?

The solution for y^{'''}-4y^{''}+y^'+6y=3e^{4t} is y=c_{1}e^{-t}+c_{2}e^{2t}+c_{3}e^{3t}+3/10 e^{4t}

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y^{'''}-4y^{''}+y^'+6y=3e^{4t}

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} + 6 y = 3 e^{4 t}

y = c_{1} e^{- t} + c_{2} e^{2 t} + c_{3} e^{3 t} + \frac{3}{10} e^{4 t}

Solution steps

y^{'''} (t) - 4 y^{''} (t) + y^{'} (t) + 6 y = 3 e^{4 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{2 t} + c_{3} e^{3 t} + \frac{3}{10} e^{4 t}

y = c_{1} e^{- t} + c_{2} e^{2 t} + c_{3} e^{3 t} + \frac{3}{10} e^{4 t}

y^{^{''}} + 4 y = sin (t) - H (t - 2 π) sin (t - 2 π)

y^{^{''}} + y = sin (a t)

y^{^{''}} - 4 y^{^{'}} + 13 y = 24 e^{2 x} cos (3 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e^{- t}

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

What is the solution for y^{'''}-4y^{''}+y^'+6y=3e^{4t} ?
The solution for y^{'''}-4y^{''}+y^'+6y=3e^{4t} is y=c_{1}e^{-t}+c_{2}e^{2t}+c_{3}e^{3t}+3/10 e^{4t}