What is the solution for y^{'''}-y^{''}-8y^'+12y=0 ?

The solution for y^{'''}-y^{''}-8y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}e^{2t}+c_{3}te^{2t}

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y^{'''}-y^{''}-8y^'+12y=0

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 0

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

Solution steps

y^{'''} (t) - y^{''} (t) - 8 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

x \frac{d y}{d x} - 2 y = 0

\int 2 x (x^{2} - 9)^{3} d x

\int_{2}^{3} \frac{21}{3 - x} d x

\int cos (π t) d t

t an g e n t y = x (27 x^{- 2} + 4)

What is the solution for y^{'''}-y^{''}-8y^'+12y=0 ?
The solution for y^{'''}-y^{''}-8y^'+12y=0 is y=c_{1}e^{-3t}+c_{2}e^{2t}+c_{3}te^{2t}