What is the solution for y^{'''}-y^{''}-16y^'+16y=0 ?

The solution for y^{'''}-y^{''}-16y^'+16y=0 is y=c_{1}e^t+c_{2}e^{-4t}+c_{3}e^{4t}

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y^{'''}-y^{''}-16y^'+16y=0

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 0

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

Solution steps

y^{'''} (t) - y^{''} (t) - 16 y^{'} (t) + 16 y = 0

Solve linear ODE:

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

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

x \to 0 lim (\frac{sin ^{5} ( x )}{x})

\frac{\partial}{\partial y} (3 x^{2} - 5 y)

y^{^{'}} = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 0

x \to a lim (9)

What is the solution for y^{'''}-y^{''}-16y^'+16y=0 ?
The solution for y^{'''}-y^{''}-16y^'+16y=0 is y=c_{1}e^t+c_{2}e^{-4t}+c_{3}e^{4t}