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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 25 y = 6 sin (5 x)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 10 x^{2} - 4 x - 15

x^{^{'''}} + 10 x^{^{''}} + 33 x^{^{'}} + 36 x = 4 t e^{- 3 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 12 e^{t} sin (t)

y^{^{''}} + 4 y = - 16 sin (2 t)

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