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

The solution for (d^2y)/(dt^2)-4(dy)/(dt)+4y=e^{-4t} is y=c_{1}e^{2t}+c_{2}te^{2t}+1/36 e^{-4t}

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(d^2y)/(dt^2)-4(dy)/(dt)+4y=e^{-4t}

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

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{36} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{36} e^{- 4 t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{1}{36} e^{- 4 t}

y^{^{''}} + 8 y^{^{'}} + 15 y = - 6 cos (5 x)

x^{^{''}} + b x^{^{'}} - c = 0

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

y^{^{''}} - y = 2 + x^{2}

y^{^{''}} + b y^{^{'}} + cy = e^{x} sin (x)

What is the solution for (d^2y)/(dt^2)-4(dy)/(dt)+4y=e^{-4t} ?
The solution for (d^2y)/(dt^2)-4(dy)/(dt)+4y=e^{-4t} is y=c_{1}e^{2t}+c_{2}te^{2t}+1/36 e^{-4t}