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}

(t (x) - t (x)^{3}) v^{'} + 2 v = 0

\frac{\partial}{\partial x} (x y e^{x - y})

\frac{7}{x ^{6}}

\frac{d}{d x} (- \frac{1}{( x + 1 ) ^{2}})

3 (1 + x^{2}) \frac{d y}{d x} = 2 x y^{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}