What is the solution for y^{''}-6y^'+8y=e^{-3t} ?

The solution for y^{''}-6y^'+8y=e^{-3t} is y=c_{1}e^{4t}+c_{2}e^{2t}+1/35 e^{-3t}

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y^{''}-6y^'+8y=e^{-3t}

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{- 3 t}

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{1}{35} e^{- 3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = e^{- 3 t}

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{1}{35} e^{- 3 t}

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{1}{35} e^{- 3 t}

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

y^{^{'''}} - y = e^{- x} + x (e^{x} + 1)

y^{^{''}} - 8 y^{^{'}} + 16 y = 3 e^{- x}

y^{^{''}} + 4 y^{^{'}} + 20 y = e^{- 2 t} sin (4 t)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 5 y = x e^{x}

What is the solution for y^{''}-6y^'+8y=e^{-3t} ?
The solution for y^{''}-6y^'+8y=e^{-3t} is y=c_{1}e^{4t}+c_{2}e^{2t}+1/35 e^{-3t}