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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 64 y = 7 e^{8 x}

y^{^{''}} + 36 y = cos (8 x)

y^{^{'''}} + y^{^{'}} = ln (x)

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

y^{^{''}} + y = sin (t) cos (t)

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