What is the solution for y^{'''}-y^{''}-4y^'+4y=0 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + y = 5 y - y^{^{''}} + 2 y^{^{'}} - y

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = 0

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

x^{^{''}} + 2 x^{^{'}} - 3 x = 0

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