What is the solution for 25y^{'''}+40y^{''}+16y^'=0 ?

The solution for 25y^{'''}+40y^{''}+16y^'=0 is y=c_{1}+c_{2}e^{-(4t)/5}+c_{3}e^{-(4t)/5}t

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25y^{'''}+40y^{''}+16y^'=0

25 y^{^{'''}} + 40 y^{^{''}} + 16 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- \frac{4 t}{5}} + c_{3} e^{- \frac{4 t}{5}} t

Solution steps

25 y^{'''} (t) + 40 y^{''} (t) + 16 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{4 t}{5}} + c_{3} e^{- \frac{4 t}{5}} t

y = c_{1} + c_{2} e^{- \frac{4 t}{5}} + c_{3} e^{- \frac{4 t}{5}} t

\frac{d}{d x} (3 arcsin (x^{3}))

\frac{d}{d x} (e^{2 x} (2 - x))

\int \frac{3 x ^{2} - 2}{x ^{2} - 4 x - 12} d x

\frac{d}{d x} (arcsin (x^{100}))

\frac{20}{2 x ^{2} + 9 x + 30}

What is the solution for 25y^{'''}+40y^{''}+16y^'=0 ?
The solution for 25y^{'''}+40y^{''}+16y^'=0 is y=c_{1}+c_{2}e^{-(4t)/5}+c_{3}e^{-(4t)/5}t