What is the solution for y^{'''}-2y^{''}-4y^'+8y=0 ?

The solution for y^{'''}-2y^{''}-4y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}e^{2t}+c_{3}te^{2t}

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y^{'''}-2y^{''}-4y^'+8y=0

y^{^{'''}} - 2 y^{^{''}} - 4 y^{^{'}} + 8 y = 0

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

Solution steps

y^{'''} (t) - 2 y^{''} (t) - 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (- 4 x^{3} + 4 x)

\int_{0}^{\frac{π}{28}} tan (14 x) d x

\frac{d}{d x} (\frac{3 x}{5 y})

\frac{m x}{b x + c}

\int (x + 5) 8 - x d x

What is the solution for y^{'''}-2y^{''}-4y^'+8y=0 ?
The solution for y^{'''}-2y^{''}-4y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}e^{2t}+c_{3}te^{2t}