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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = - 1

\frac{d ^{2} x}{d t ^{2}} + 0.75 \frac{d x}{d t} + 0.3 x = 0

y^{^{''}} + 10 y^{^{'}} + 2 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - \frac{5}{3}

x^{^{''}} + 5 x^{^{'}} + 3 x = 0, x (0) = 5, x^{^{'}} (0) = 7

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