What is the solution for (d^2y)/(dt^2)+5(dy)/(dt)+4y=2t ?

The solution for (d^2y)/(dt^2)+5(dy)/(dt)+4y=2t is y=c_{1}e^{-t}+c_{2}e^{-4t}+t/2-5/8

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(d^2y)/(dt^2)+5(dy)/(dt)+4y=2t

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 2 t

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t}{2} - \frac{5}{8}

Solution steps

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 2 t

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t}{2} - \frac{5}{8}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{t}{2} - \frac{5}{8}

- y^{^{''}} + 2 y^{^{'}} + 3 y = 1, y (0) = 0, y (1) = 0

y^{^{''}} + 4 y^{^{'}} + 8 = 0

y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{- 2 x}

y^{^{''}} + 8 y = 8 x

y^{^{'''}} + 12 y^{^{''}} + 48 y^{^{'}} + 64 y = 8 e^{- 4 x}

What is the solution for (d^2y)/(dt^2)+5(dy)/(dt)+4y=2t ?
The solution for (d^2y)/(dt^2)+5(dy)/(dt)+4y=2t is y=c_{1}e^{-t}+c_{2}e^{-4t}+t/2-5/8