What is the solution for y^{'''}-9y^{''}+28y^'-20y=0 ?

The solution for y^{'''}-9y^{''}+28y^'-20y=0 is y=c_{1}e^t+e^{4t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}-9y^{''}+28y^'-20y=0

y^{^{'''}} - 9 y^{^{''}} + 28 y^{^{'}} - 20 y = 0

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

Solution steps

y^{'''} (t) - 9 y^{''} (t) + 28 y^{'} (t) - 20 y = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

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

\frac{d ^{2} y}{d x ^{2}} - k^{2} y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

What is the solution for y^{'''}-9y^{''}+28y^'-20y=0 ?
The solution for y^{'''}-9y^{''}+28y^'-20y=0 is y=c_{1}e^t+e^{4t}(c_{2}cos(2t)+c_{3}sin(2t))