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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

g (x) = x^{\frac{5}{2}}

\int_{0}^{\infty} \frac{1}{4 + x ^{2}} d x

\frac{d}{d x} (e^{x} + x)

\frac{d}{d x} (9 x^{2} cos (x) cot (x))

\int_{1}^{2} \frac{1}{3 - x} d x

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