What is the solution for y^{''''}-5y^{''}+4y=0 ?

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

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y^{''''}-5y^{''}+4y=0

y^{^{''''}} - 5 y^{^{''}} + 4 y = 0

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

Solution steps

y^{(4)} (t) - 5 y^{''} (t) + 4 y = 0

Solve linear ODE:

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

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

n = 1 \sum \infty arctan (11 n)

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

f (x) = x^{2}, (1, - 3)

\frac{\partial}{\partial x} (x^{2} sin (k x))

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

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