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

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

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

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

y = \frac{e ^{- t} t ^{2}}{2} - e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{- t} t ^{2}}{2} - e^{- t}

y = \frac{e ^{- t} t ^{2}}{2} - e^{- t}

\frac{d}{d x} (\frac{60}{x ^{4}})

x \to 2 lim (\frac{x ^{2} - 1}{x - 1})

n = 0 \sum \infty arctan (n)

\int_{- 3}^{3} 9 - x^{2} d x

y^{3} = \frac{d y}{d x} + y

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