What is the solution for y^{'''}-5y^{''}-2y^'+24y=0 ?

The solution for y^{'''}-5y^{''}-2y^'+24y=0 is y=c_{1}e^{-2t}+c_{2}e^{3t}+c_{3}e^{4t}

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y^{'''}-5y^{''}-2y^'+24y=0

y^{^{'''}} - 5 y^{^{''}} - 2 y^{^{'}} + 24 y = 0

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

Solution steps

y^{'''} (t) - 5 y^{''} (t) - 2 y^{'} (t) + 24 y = 0

Solve linear ODE:

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

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

f^{^{''}} (x) + f (x) = 0, f (0) = 1

f^{^{''}} (x) + f (x) = 0, f (0) = 0

\frac{d ^{2} x}{d t ^{2}} - 5 \frac{d x}{d t} + 4 x = 0

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 4 y = 0

20 x^{^{''}} + 9 x^{^{'}} + x = 0, x (0) = 0, x^{^{'}} (0) = 4

What is the solution for y^{'''}-5y^{''}-2y^'+24y=0 ?
The solution for y^{'''}-5y^{''}-2y^'+24y=0 is y=c_{1}e^{-2t}+c_{2}e^{3t}+c_{3}e^{4t}