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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} - 8 y + 17 y = 0

y^{^{''}} + 14 y^{^{'}} + 48 y = 0, y (0) = 9, y^{^{'}} (0) = - 66

y^{^{''''}} = 625 y

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