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

The solution for y^{''}+4y^'+3y=6te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(e^{3t}t)/4-5/48 e^{3t}

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y^{''}+4y^'+3y=6te^{3t}

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

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{3 t} t}{4} - \frac{5}{48} e^{3 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{3 t} t}{4} - \frac{5}{48} e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{3 t} t}{4} - \frac{5}{48} e^{3 t}

y^{^{''}} - 9 y^{^{'}} + 14 y = 30 t - 34

y^{^{''}} + y = 6 sin (2 t), y (0) = 3, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} + 4 y = x

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

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

What is the solution for y^{''}+4y^'+3y=6te^{3t} ?
The solution for y^{''}+4y^'+3y=6te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(e^{3t}t)/4-5/48 e^{3t}