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

The solution for y^{''}-4y^'+4y=e^t is y=c_{1}e^{2t}+c_{2}te^{2t}+e^t

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

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

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + e^{t}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + e^{t}

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 2 e^{t}

y^{^{''}} + 4 y = 3 x^{3}

x^{^{''}} + 2 x^{^{'}} + x = e^{- t}

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 t)

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 e^{2 x}

What is the solution for y^{''}-4y^'+4y=e^t ?
The solution for y^{''}-4y^'+4y=e^t is y=c_{1}e^{2t}+c_{2}te^{2t}+e^t