What is the solution for y^{''}-4y^'+4y=e^{-t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'+4y=e^{-t},y(0)=0,y^'(0)=0 is y=-1/9 e^{2t}+(e^{2t}t)/3+1/9 e^{-t}

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y^{''}-4y^'+4y=e^{-t},y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{9} e^{2 t} + \frac{e ^{2 t} t}{3} + \frac{1}{9} e^{- t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = e^{- t}

Solve linear ODE:

y = - \frac{1}{9} e^{2 t} + \frac{e ^{2 t} t}{3} + \frac{1}{9} e^{- t}

y = - \frac{1}{9} e^{2 t} + \frac{e ^{2 t} t}{3} + \frac{1}{9} e^{- t}

What is the solution for y^{''}-4y^'+4y=e^{-t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'+4y=e^{-t},y(0)=0,y^'(0)=0 is y=-1/9 e^{2t}+(e^{2t}t)/3+1/9 e^{-t}