What is the solution for 49y^{'''}+28y^{''}+4y^'=0 ?

The solution for 49y^{'''}+28y^{''}+4y^'=0 is y=c_{1}+c_{2}e^{-(2t)/7}+c_{3}e^{-(2t)/7}t

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49y^{'''}+28y^{''}+4y^'=0

49 y^{^{'''}} + 28 y^{^{''}} + 4 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- \frac{2 t}{7}} + c_{3} e^{- \frac{2 t}{7}} t

Solution steps

49 y^{'''} (t) + 28 y^{''} (t) + 4 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{2 t}{7}} + c_{3} e^{- \frac{2 t}{7}} t

y = c_{1} + c_{2} e^{- \frac{2 t}{7}} + c_{3} e^{- \frac{2 t}{7}} t

y^{^{'''}} - 5 y^{^{''}} - 36 y^{^{'}} = 0

\frac{3}{2} y^{^{''}} + 3 y^{^{'}} + 5 y = 0

56 y^{^{''}} + 33 y^{^{'}} - 14 y = 0

(d^{3} - 2 d^{2} - d + 2) y = 0

4 y^{^{'''}} + 17 y^{^{''}} - 16 y^{^{'}} - 5 y = 0

What is the solution for 49y^{'''}+28y^{''}+4y^'=0 ?
The solution for 49y^{'''}+28y^{''}+4y^'=0 is y=c_{1}+c_{2}e^{-(2t)/7}+c_{3}e^{-(2t)/7}t