What is the solution for 25y^{''}+10y^'+1y=0 ?

The solution for 25y^{''}+10y^'+1y=0 is y=c_{1}e^{-t/5}+c_{2}e^{-t/5}t

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25y^{''}+10y^'+1y=0

25 y^{^{''}} + 10 y^{^{'}} + 1 y = 0

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

Solution steps

25 y^{''} (t) + 10 y^{'} (t) + 1 y = 0

Solve linear ODE:

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

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

3 y^{^{''}} - 13 y^{^{'}} + 14 y = 0

2 y^{^{''}} + 13 y^{^{'}} - 7 y = 0

10 x^{^{''}} + 7 x^{^{'}} + x = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 5

4 y^{^{''}} - y^{^{'}} - 6 y = 0

What is the solution for 25y^{''}+10y^'+1y=0 ?
The solution for 25y^{''}+10y^'+1y=0 is y=c_{1}e^{-t/5}+c_{2}e^{-t/5}t