What is the solution for y^{'''}-3y^{''}-4y^'+12y=0 ?

The solution for y^{'''}-3y^{''}-4y^'+12y=0 is y=c_{1}e^{3t}+c_{2}e^{-2t}+c_{3}e^{2t}

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y^{'''}-3y^{''}-4y^'+12y=0

y^{^{'''}} - 3 y^{^{''}} - 4 y^{^{'}} + 12 y = 0

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

Solution steps

y^{'''} (t) - 3 y^{''} (t) - 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

y^{^{''}} + y = 0, y (0) = sin (x)

9 \frac{d ^{2} y}{d t ^{2}} - 24 \frac{d y}{d t} + 16 y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

(4 d^{4} - 8 d^{3} - 4 d^{2} + 8 d) y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{'''}-3y^{''}-4y^'+12y=0 ?
The solution for y^{'''}-3y^{''}-4y^'+12y=0 is y=c_{1}e^{3t}+c_{2}e^{-2t}+c_{3}e^{2t}