What is the solution for 25y^{''}+20y^'+4y=0 ?

The solution for 25y^{''}+20y^'+4y=0 is y=c_{1}e^{-(2t)/5}+c_{2}e^{-(2t)/5}t

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25y^{''}+20y^'+4y=0

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 0

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

Solution steps

25 y^{''} (t) + 20 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

4 \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 10 y = 0

(4 D^{2} - 4 D + 5) y = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 17 y = 0, y (0) = 2, y^{^{'}} (0) = 10

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = α, y^{^{'}} (0) = 2

What is the solution for 25y^{''}+20y^'+4y=0 ?
The solution for 25y^{''}+20y^'+4y=0 is y=c_{1}e^{-(2t)/5}+c_{2}e^{-(2t)/5}t