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

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

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

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

y = - \frac{1}{2} e^{- 2 t} - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{2} e^{- 2 t} - \frac{1}{2}

y = - \frac{1}{2} e^{- 2 t} - \frac{1}{2}

\frac{d ^{2} y}{d t ^{2}} + 2 y = cos (a t)

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

y^{^{''}} - y^{^{'}} - 30 y = 84 e^{2 t}

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

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

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