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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 9 y^{^{''}} + 108 y = 0

y^{^{''}} + y^{^{'}} + 56 y = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = \frac{53}{5}

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = \frac{3}{5}, y^{^{'}} (0) = - \frac{1}{5}

\frac{d ^{4} y}{d x ^{4}} + 8 \frac{d ^{2} y}{d x ^{2}} + 6 y = 0

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