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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = x

\int \frac{4 x ^{3} + 2}{x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (sin (5 x))

y = cot (1 - 2 x^{2})

x y^{^{'}} + 3 y = \frac{2 e ^{2 x}}{x ^{2}}

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