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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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\frac{\partial}{\partial y} (2 e^{x} ln (y))

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