What is the solution for y^{''}+5y^'+6y=te^{-t} ?

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

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y^{''}+5y^'+6y=te^{-t}

y^{^{''}} + 5 y^{^{'}} + 6 y = t e^{- t}

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

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = t e^{- t}

Solve linear ODE:

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

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

x \to 3 lim (9 - x^{2})

\frac{d y}{d x} = 2 x y - y, y (0) = 2

\int \frac{1}{sin ^{2} ( t ) - 1} d t

\int 2 cos (x) + sec^{2} (x) d x

\frac{3 s + 3}{( s ^{2} - 9 )}

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