What is the solution for 25y^{''}+80y^'+64y=0 ?

The solution for 25y^{''}+80y^'+64y=0 is y=c_{1}e^{-(8t)/5}+c_{2}e^{-(8t)/5}t

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25y^{''}+80y^'+64y=0

25 y^{^{''}} + 80 y^{^{'}} + 64 y = 0

y = c_{1} e^{- \frac{8 t}{5}} + c_{2} e^{- \frac{8 t}{5}} t

Solution steps

25 y^{''} (t) + 80 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{8 t}{5}} + c_{2} e^{- \frac{8 t}{5}} t

y = c_{1} e^{- \frac{8 t}{5}} + c_{2} e^{- \frac{8 t}{5}} t

\int x^{3} 169 - x^{2} d x

\frac{d}{d x} (\frac{2}{2 x})

\frac{d}{d x} (2 sin (x) - 3)

y = \frac{6}{( x - 1 ) ^{2}}, at

\frac{d}{d x} ((sec (4 x) + tan (2 x))^{5})

What is the solution for 25y^{''}+80y^'+64y=0 ?
The solution for 25y^{''}+80y^'+64y=0 is y=c_{1}e^{-(8t)/5}+c_{2}e^{-(8t)/5}t