What is the solution for y^{'''}-5y^{''}+24y^'-20y=0 ?

The solution for y^{'''}-5y^{''}+24y^'-20y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(4t)+c_{3}sin(4t))

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y^{'''}-5y^{''}+24y^'-20y=0

y^{^{'''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0

y = c_{1} e^{t} + e^{2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

Solution steps

y^{'''} (t) - 5 y^{''} (t) + 24 y^{'} (t) - 20 y = 0

Solve linear ODE:

y = c_{1} e^{t} + e^{2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

y = c_{1} e^{t} + e^{2 t} (c_{2} cos (4 t) + c_{3} sin (4 t))

y^{^{''}} - 7 y = 0

(D^{5} - 15 D^{3} + 10 D^{2} + 60 D - 72) y = 0

y^{^{'''}} - 7 y^{^{''}} - 49 y^{^{'}} + 343 y = 0

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} + 12 y = 0

y^{^{''}} + 12 y^{^{'}} + 37 y = 0

What is the solution for y^{'''}-5y^{''}+24y^'-20y=0 ?
The solution for y^{'''}-5y^{''}+24y^'-20y=0 is y=c_{1}e^t+e^{2t}(c_{2}cos(4t)+c_{3}sin(4t))