What is the solution for y^{''}+5y^'+4y=2te^{3t} ?

The solution for y^{''}+5y^'+4y=2te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{3t}t)/(14)-11/392 e^{3t}

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y^{''}+5y^'+4y=2te^{3t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 t e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{3 t} t}{14} - \frac{11}{392} e^{3 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = 2 t e^{3 t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{3 t} t}{14} - \frac{11}{392} e^{3 t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{3 t} t}{14} - \frac{11}{392} e^{3 t}

\int_{0}^{6} 36 - x^{2} d x

x \to 0 + lim (ln (ln (x)))

\frac{d y}{d x} = \frac{1}{sec ^{2} ( y )}

\frac{d}{d x} (\frac{x ^{4} + 2}{ln ( x )})

\int \frac{x}{e ^{4 x}} d x

What is the solution for y^{''}+5y^'+4y=2te^{3t} ?
The solution for y^{''}+5y^'+4y=2te^{3t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{3t}t)/(14)-11/392 e^{3t}