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+7

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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 + 7

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 + 7

y = c_{1} e^{t} + c_{2} t e^{t} + t^{2} + 4 t + 7

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 cos (4 t)

x^{^{''}} + 4 x = - 6 sin (2 t) + 3 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 = x

y^{^{''}} + y^{^{'}} = \frac{1}{sin ( x )}

y^{^{''}} + 16 y = 20 sin (4 t), y (0) = 0, y^{^{'}} (0) = 0

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+7