What is the solution for x^{''}-4x^'+4x=4e^{2t} ?

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

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x^{''}-4x^'+4x=4e^{2t}

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}

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

Solution steps

x^{''} (t) - 4 x^{'} (t) + 4 x = 4 e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} - 4 y = 2 e^{4 t}

y^{^{''}} - 6 y^{^{'}} + 9 y = 25 e^{t} sin (t)

(D^{2} + 7 D + 6) y = e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = ln (x)

y^{^{''}} = 2 y + 18 tan^{3} (x)

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