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

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

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

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

y = c_{1} e^{4 t} + c_{2} + c_{3} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} + c_{3} t

y = c_{1} e^{4 t} + c_{2} + c_{3} t

y^{^{''}} + 2 y^{^{'}} + 17 y = 0, y (0) = 4, y^{^{'}} (0) = - 3

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

(4 D^{4} + 20 D^{3} + 35 D^{2} + 25 D + 6) y = 0

(D^{2} - 4 D + 4) y = 0

25 x^{^{''}} - 20 x^{^{'}} + 4 x = 0

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