What is the solution for y^{''}-2y^'+y=8e^t ?

The solution for y^{''}-2y^'+y=8e^t is y=c_{1}e^t+c_{2}te^t+4t^2e^t

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

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

y = c_{1} e^{t} + c_{2} t e^{t} + 4 t^{2} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + 4 t^{2} e^{t}

y = c_{1} e^{t} + c_{2} t e^{t} + 4 t^{2} e^{t}

y^{^{''}} + y = 2 sin (2 t), y (0) = 6, y^{^{'}} (0) = 0

d, d (t) = 12 t + 20

y^{^{''}} + 2 y^{^{'}} + y = 12 x e^{- x}

2 y^{^{''}} + 5 y^{^{'}} + 4 x = 0

y^{^{''}} - y^{^{'}} = 3 x^{2}

What is the solution for y^{''}-2y^'+y=8e^t ?
The solution for y^{''}-2y^'+y=8e^t is y=c_{1}e^t+c_{2}te^t+4t^2e^t