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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0

12 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

(D^{6} - 4 D^{4} + 4 D^{2}) y = 0

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} = 16 y

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