What is the solution for y^{''}+5y^'+6y=2e^{-t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+5y^'+6y=2e^{-t},y(0)=0,y^'(0)=0 is y=-2e^{-2t}+e^{-3t}+e^{-t}

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y^{''}+5y^'+6y=2e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = - 2 e^{- 2 t} + e^{- 3 t} + e^{- t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 2 e^{- t}

Solve linear ODE:

y = - 2 e^{- 2 t} + e^{- 3 t} + e^{- t}

y = - 2 e^{- 2 t} + e^{- 3 t} + e^{- t}

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 4 x^{2} + 8 x + 10

(D^{3} + D - 10) y = 29 e^{4 x}

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

y^{^{''''}} - 4 y^{^{''}} = x^{2}

y^{^{''}} + 2 y^{^{'}} + 5 y = 12 e^{- t} cos (2 t)

What is the solution for y^{''}+5y^'+6y=2e^{-t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+5y^'+6y=2e^{-t},y(0)=0,y^'(0)=0 is y=-2e^{-2t}+e^{-3t}+e^{-t}