What is the solution for y^{''''}-2y^{''}+1=0 ?

The solution for y^{''''}-2y^{''}+1=0 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}+c_{3}+c_{4}t+(t^2)/4

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y^{''''}-2y^{''}+1=0

y^{^{''''}} - 2 y^{^{''}} + 1 = 0

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t + \frac{t ^{2}}{4}

Solution steps

y^{(4)} (t) - 2 y^{''} (t) + 1 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t + \frac{t ^{2}}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + c_{3} + c_{4} t + \frac{t ^{2}}{4}

x^{^{''}} + b \cdot x^{^{'}} + a = 0

x^{^{''}} - 6 x - 16 x = sin (2 t)

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

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

y^{^{''}} + 6 y^{^{'}} + 3 x = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''''}-2y^{''}+1=0 ?
The solution for y^{''''}-2y^{''}+1=0 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}+c_{3}+c_{4}t+(t^2)/4