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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} )} + 3 e^{t}

y^{^{''}} + 10 y^{^{'}} + 20 y = 2695 e^{2 t} cos (11 t)

(D^{2} + 4) y = cos (2 x)

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