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

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

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

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

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t} + 2 t

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t} + 2 t

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t} + 2 t

(\frac{d ^{2} y}{d x ^{2}} - 9 y) = e^{- 4 x} cos (e^{- x})

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

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

\frac{d ^{2} y}{d x ^{2}} + 36 y = 288 sin (6 x) + 96 cos (6 x)

y^{^{''}} + y = sec (x) + tan (x)

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