What is the solution for 144y^{'''}+120y^{''}+25y^'=0 ?

The solution for 144y^{'''}+120y^{''}+25y^'=0 is y=c_{1}+c_{2}e^{-(5t)/(12)}+c_{3}e^{-(5t)/(12)}t

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144y^{'''}+120y^{''}+25y^'=0

144 y^{^{'''}} + 120 y^{^{''}} + 25 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- \frac{5 t}{12}} + c_{3} e^{- \frac{5 t}{12}} t

Solution steps

144 y^{'''} (t) + 120 y^{''} (t) + 25 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{5 t}{12}} + c_{3} e^{- \frac{5 t}{12}} t

y = c_{1} + c_{2} e^{- \frac{5 t}{12}} + c_{3} e^{- \frac{5 t}{12}} t

\int (4 x + 7) e^{x} d x

\int 30 x^{29} d x

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 2}{2 x - 5})

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0, y (5) = 0, y^{^{'}} (0) = - e^{2}

x = y^{2} + 1, x = y + 3

What is the solution for 144y^{'''}+120y^{''}+25y^'=0 ?
The solution for 144y^{'''}+120y^{''}+25y^'=0 is y=c_{1}+c_{2}e^{-(5t)/(12)}+c_{3}e^{-(5t)/(12)}t