What is the solution for y^{'''}-5y^{''}-22y^'+56y=0 ?

The solution for y^{'''}-5y^{''}-22y^'+56y=0 is y=c_{1}e^{2t}+c_{2}e^{-4t}+c_{3}e^{7t}

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y^{'''}-5y^{''}-22y^'+56y=0

y^{^{'''}} - 5 y^{^{''}} - 22 y^{^{'}} + 56 y = 0

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + c_{3} e^{7 t}

Solution steps

y^{'''} (t) - 5 y^{''} (t) - 22 y^{'} (t) + 56 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + c_{3} e^{7 t}

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} + c_{3} e^{7 t}

y = (x^{3} - 2) (x^{2} - 3 x + 1), (2, - 6)

\frac{d x}{d t} = 7 x

4 x + 3 x^{2}

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

\int 2 x (x^{2} + 3)^{- 4} d x

What is the solution for y^{'''}-5y^{''}-22y^'+56y=0 ?
The solution for y^{'''}-5y^{''}-22y^'+56y=0 is y=c_{1}e^{2t}+c_{2}e^{-4t}+c_{3}e^{7t}