What is the solution for y^{'''}-7y^{''}+25y^'-39y=0 ?

The solution for y^{'''}-7y^{''}+25y^'-39y=0 is y=c_{1}e^{3t}+e^{2t}(c_{2}cos(3t)+c_{3}sin(3t))

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y^{'''}-7y^{''}+25y^'-39y=0

y^{^{'''}} - 7 y^{^{''}} + 25 y^{^{'}} - 39 y = 0

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

Solution steps

y^{'''} (t) - 7 y^{''} (t) + 25 y^{'} (t) - 39 y = 0

Solve linear ODE:

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

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

2 \cdot 0.04 \cdot x^{^{'}} + 0.25 \cdot 0.04 \cdot x^{^{''}} + x = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 0, y^{^{'}} (0) = 1

x^{^{''}} + 16.33 x = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 1, y (1) = 0

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{'''}-7y^{''}+25y^'-39y=0 ?
The solution for y^{'''}-7y^{''}+25y^'-39y=0 is y=c_{1}e^{3t}+e^{2t}(c_{2}cos(3t)+c_{3}sin(3t))