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

The solution for y^{''}-2y^'+y=t^2-1 is y=c_{1}e^t+c_{2}te^t+t^2+4t+5

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y^{''}-2y^'+y=t^2-1

y^{^{''}} - 2 y^{^{'}} + y = t^{2} - 1

y = c_{1} e^{t} + c_{2} t e^{t} + t^{2} + 4 t + 5

Solution steps

y^{''} (t) - 2 y^{'} (t) + y = t^{2} - 1

Solve linear ODE:

y = c_{1} e^{t} + c_{2} t e^{t} + t^{2} + 4 t + 5

y = c_{1} e^{t} + c_{2} t e^{t} + t^{2} + 4 t + 5

y^{^{''}} - 2 y^{^{'}} - 3 y = cos (2 x)

x^{^{''}} + 4 x = sin (5 t)

y^{^{''}} + y = sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = tan (3 t)

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 e^{x} - 18 e^{- x}

What is the solution for y^{''}-2y^'+y=t^2-1 ?
The solution for y^{''}-2y^'+y=t^2-1 is y=c_{1}e^t+c_{2}te^t+t^2+4t+5