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}

y^{^{'}} - x e^{y} = 2 e^{y}

\frac{d y}{d x} + 5 y = 3

\int \frac{3 x - 4}{x ^{2} - 3 x + 2} d x

x + y^{2} = 42, x + y = 0

x = 25 - y^{2}, x = y^{2} - 25

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}