What is the solution for y^{'''}-3y^{''}+4y^'-12y=0 ?

The solution for y^{'''}-3y^{''}+4y^'-12y=0 is y=c_{1}e^{3t}+c_{2}cos(2t)+c_{3}sin(2t)

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y^{'''}-3y^{''}+4y^'-12y=0

y^{^{'''}} - 3 y^{^{''}} + 4 y^{^{'}} - 12 y = 0

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

Solution steps

y^{'''} (t) - 3 y^{''} (t) + 4 y^{'} (t) - 12 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = - 2, y^{^{'}} (0) = 0

\frac{d ^{2}}{d θ ^{2}} (y (θ)) + y (θ) = 0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0

y^{^{'''}} - 3 y^{^{''}} - 4 y^{^{'}} + 12 y = 0

y^{^{''}} + y = 0, y (0) = sin (x)

What is the solution for y^{'''}-3y^{''}+4y^'-12y=0 ?
The solution for y^{'''}-3y^{''}+4y^'-12y=0 is y=c_{1}e^{3t}+c_{2}cos(2t)+c_{3}sin(2t)