What is the solution for y^{''}-3y^'+2y=e^{-4t},y(0)=1 ?

The solution for y^{''}-3y^'+2y=e^{-4t},y(0)=1 is y=(-c_{2}+29/30)e^{2t}+c_{2}e^t+1/30 e^{-4t}

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y^{''}-3y^'+2y=e^{-4t},y(0)=1

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1

y = (- c_{2} + \frac{29}{30}) e^{2 t} + c_{2} e^{t} + \frac{1}{30} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = (- c_{2} + \frac{29}{30}) e^{2 t} + c_{2} e^{t} + \frac{1}{30} e^{- 4 t}

y = (- c_{2} + \frac{29}{30}) e^{2 t} + c_{2} e^{t} + \frac{1}{30} e^{- 4 t}

y^{^{''}} + 3 y^{^{'}} = 2 x - 3

y^{^{''}} + 2 y^{^{'}} + y = 3 x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 = 0

y^{^{'''}} - y = 1

y^{^{''}} + 7 y^{^{'}} + 12 y = 4 e^{5 t}, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{''}-3y^'+2y=e^{-4t},y(0)=1 ?
The solution for y^{''}-3y^'+2y=e^{-4t},y(0)=1 is y=(-c_{2}+29/30)e^{2t}+c_{2}e^t+1/30 e^{-4t}